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<h2 id="eulers-approach">Euler's approach</h2> <p>Euler's original derivation of the value π 2 6 {\textstyle {\frac {\pi ^{2}}{6}}} essentially extended observations about finite <a href="/facts/Polynomial/Lzak8VVx">polynomials</a> and assumed that these same properties hold true for infinite series. </p><p>Of course, Euler's original reasoning requires justification (100 years later, <a href="/facts/Karl_Weierstrass/4U8vdQzL">Karl Weierstrass</a> proved that Euler's representation of the sine function as an infinite product is valid, by the <a href="/facts/Weierstrass_factorization_theorem/atdInY5m">Weierstrass factorization theorem</a>), but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community. </p><p>To follow Euler's argument, recall the <a href="/facts/Taylor_series/M0aTuftH">Taylor series</a> expansion of the <a href="/facts/Trigonometric_function/czej7ur0">sine function</a> sin x = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + ⋯ {\displaystyle \sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots } Dividing through by x gives sin x x = 1 − x 2 3 ! + x 4 5 ! − x 6 7 ! + ⋯ . {\displaystyle {\frac {\sin x}{x}}=1-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}-{\frac {x^{6}}{7!}}+\cdots .} </p><p>The <a href="/facts/Weierstrass_factorization_theorem/atdInY5m">Weierstrass factorization theorem</a> shows that the right-hand side is the product of linear factors given by its roots, just as for finite polynomials. Euler assumed this as a <a href="/facts/Heuristic/BcBEkv9c">heuristic</a> for expanding an infinite degree <a href="/facts/Polynomial/Lzak8VVx">polynomial</a> in terms of its roots, but in fact it is not always true for general P ( x ) {\displaystyle P(x)} .<a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a> This factorization expands the equation into: sin x x = ( 1 − x π ) ( 1 + x π ) ( 1 − x 2 π ) ( 1 + x 2 π ) ( 1 − x 3 π ) ( 1 + x 3 π ) ⋯ = ( 1 − x 2 π 2 ) ( 1 − x 2 4 π 2 ) ( 1 − x 2 9 π 2 ) ⋯ {\displaystyle {\begin{aligned}{\frac {\sin x}{x}}&=\left(1-{\frac {x}{\pi }}\right)\left(1+{\frac {x}{\pi }}\right)\left(1-{\frac {x}{2\pi }}\right)\left(1+{\frac {x}{2\pi }}\right)\left(1-{\frac {x}{3\pi }}\right)\left(1+{\frac {x}{3\pi }}\right)\cdots \\&=\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots \end{aligned}}} </p><p>If we formally multiply out this product and collect all the <i>x</i>2 terms (we are allowed to do so because of <a href="/facts/Newton%2527s_identities/5J9flXxA">Newton's identities</a>), we see by induction that the <i>x</i>2 coefficient of sin <i>x</i>/<i>x</i> is <a class="footnote-ref" id="fnref:6" href="#fn:6"><sup>6</sup></a> − ( 1 π 2 + 1 4 π 2 + 1 9 π 2 + ⋯ ) = − 1 π 2 ∑ n = 1 ∞ 1 n 2 . {\displaystyle -\left({\frac {1}{\pi ^{2}}}+{\frac {1}{4\pi ^{2}}}+{\frac {1}{9\pi ^{2}}}+\cdots \right)=-{\frac {1}{\pi ^{2}}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}.} </p><p>But from the original infinite series expansion of sin <i>x</i>/<i>x</i>, the coefficient of <i>x</i>2 is −1/3! = −1/6. These two coefficients must be equal; thus, − 1 6 = − 1 π 2 ∑ n = 1 ∞ 1 n 2 . {\displaystyle -{\frac {1}{6}}=-{\frac {1}{\pi ^{2}}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}.} </p><p>Multiplying both sides of this equation by −π2 gives the sum of the reciprocals of the positive square integers.<a class="footnote-ref" id="fnref:7" href="#fn:7"><sup>7</sup></a> ∑ n = 1 ∞ 1 n 2 = π 2 6 . {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}.} </p> <h3>Generalizations of Euler's method using elementary symmetric polynomials</h3> <p>Using formulae obtained from <a href="/facts/Elementary_symmetric_polynomial/9RFXe9b6">elementary symmetric polynomials</a>,<a class="footnote-ref" id="fnref:8" href="#fn:8"><sup>8</sup></a> this same approach can be used to enumerate formulae for the even-indexed <a href="/facts/Zeta_constants/3W4ClRUP">even zeta constants</a> which have the following known formula expanded by the <a href="/facts/Bernoulli_numbers/RH0mblhs">Bernoulli numbers</a>: ζ ( 2 n ) = ( − 1 ) n − 1 ( 2 π ) 2 n 2 ⋅ ( 2 n ) ! B 2 n . {\displaystyle \zeta (2n)={\frac {(-1)^{n-1}(2\pi )^{2n}}{2\cdot (2n)!}}B_{2n}.} </p><p>For example, let the partial product for sin ( x ) {\displaystyle \sin(x)} expanded as above be defined by S n ( x ) x := ∏ k = 1 n ( 1 − x 2 k 2 ⋅ π 2 ) {\displaystyle {\frac {S_{n}(x)}{x}}:=\prod \limits _{k=1}^{n}\left(1-{\frac {x^{2}}{k^{2}\cdot \pi ^{2}}}\right)} . Then using known <a href="/facts/Newton%2527s_identities/5J9flXxA">formulas for elementary symmetric polynomials</a> (a.k.a., Newton's formulas expanded in terms of <a href="/facts/Power_sum/8Dxbu6Bd">power sum</a> identities), we can see (for example) that [ x 4 ] S n ( x ) x = 1 2 π 4 ( ( H n ( 2 ) ) 2 − H n ( 4 ) ) → n → ∞ 1 2 π 4 ( ζ ( 2 ) 2 − ζ ( 4 ) ) ⟹ ζ ( 4 ) = π 4 90 = − 2 π 4 ⋅ [ x 4 ] sin ( x ) x + π 4 36 [ x 6 ] S n ( x ) x = − 1 6 π 6 ( ( H n ( 2 ) ) 3 − 2 H n ( 2 ) H n ( 4 ) + 2 H n ( 6 ) ) → n → ∞ 1 6 π 6 ( ζ ( 2 ) 3 − 3 ζ ( 2 ) ζ ( 4 ) + 2 ζ ( 6 ) ) ⟹ ζ ( 6 ) = π 6 945 = − 3 ⋅ π 6 [ x 6 ] sin ( x ) x − 2 3 π 2 6 π 4 90 + π 6 216 , {\displaystyle {\begin{aligned}\left[x^{4}\right]{\frac {S_{n}(x)}{x}}&={\frac {1}{2\pi ^{4}}}\left(\left(H_{n}^{(2)}\right)^{2}-H_{n}^{(4)}\right)\qquad \xrightarrow {n\rightarrow \infty } \qquad {\frac {1}{2\pi ^{4}}}\left(\zeta (2)^{2}-\zeta (4)\right)\\[4pt]&\qquad \implies \zeta (4)={\frac {\pi ^{4}}{90}}=-2\pi ^{4}\cdot [x^{4}]{\frac {\sin(x)}{x}}+{\frac {\pi ^{4}}{36}}\\[8pt]\left[x^{6}\right]{\frac {S_{n}(x)}{x}}&=-{\frac {1}{6\pi ^{6}}}\left(\left(H_{n}^{(2)}\right)^{3}-2H_{n}^{(2)}H_{n}^{(4)}+2H_{n}^{(6)}\right)\qquad \xrightarrow {n\rightarrow \infty } \qquad {\frac {1}{6\pi ^{6}}}\left(\zeta (2)^{3}-3\zeta (2)\zeta (4)+2\zeta (6)\right)\\[4pt]&\qquad \implies \zeta (6)={\frac {\pi ^{6}}{945}}=-3\cdot \pi ^{6}[x^{6}]{\frac {\sin(x)}{x}}-{\frac {2}{3}}{\frac {\pi ^{2}}{6}}{\frac {\pi ^{4}}{90}}+{\frac {\pi ^{6}}{216}},\end{aligned}}} </p><p>and so on for subsequent coefficients of [ x 2 k ] S n ( x ) x {\displaystyle [x^{2k}]{\frac {S_{n}(x)}{x}}} . There are <a href="/facts/Newton%2527s_identities/5J9flXxA">other forms of Newton's identities</a> expressing the (finite) power sums H n ( 2 k ) {\displaystyle H_{n}^{(2k)}} in terms of the <a href="/facts/Elementary_symmetric_polynomial/9RFXe9b6">elementary symmetric polynomials</a>, e i ≡ e i ( − π 2 1 2 , − π 2 2 2 , − π 2 3 2 , − π 2 4 2 , … ) , {\displaystyle e_{i}\equiv e_{i}\left(-{\frac {\pi ^{2}}{1^{2}}},-{\frac {\pi ^{2}}{2^{2}}},-{\frac {\pi ^{2}}{3^{2}}},-{\frac {\pi ^{2}}{4^{2}}},\ldots \right),} but we can go a more direct route to expressing non-recursive formulas for ζ ( 2 k ) {\displaystyle \zeta (2k)} using the method of <a href="/facts/Elementary_symmetric_polynomial/9RFXe9b6">elementary symmetric polynomials</a>. Namely, we have a recurrence relation between the elementary symmetric polynomials and the <a href="/facts/Power_sum_symmetric_polynomial/bmMAz7GG">power sum polynomials</a> given as on <a href="/facts/Newton%2527s_identities/5J9flXxA">this page</a> by ( − 1 ) k k e k ( x 1 , … , x n ) = ∑ j = 1 k ( − 1 ) k − j − 1 p j ( x 1 , … , x n ) e k − j ( x 1 , … , x n ) , {\displaystyle (-1)^{k}ke_{k}(x_{1},\ldots ,x_{n})=\sum _{j=1}^{k}(-1)^{k-j-1}p_{j}(x_{1},\ldots ,x_{n})e_{k-j}(x_{1},\ldots ,x_{n}),} </p><p>which in our situation equates to the limiting recurrence relation (or <a href="/facts/Generating_function/K7UVjnjL">generating function</a> convolution, or <a href="/facts/Cauchy_product/UF8fbLwg">product</a>) expanded as π 2 k 2 ⋅ ( 2 k ) ⋅ ( − 1 ) k ( 2 k + 1 ) ! = − [ x 2 k ] sin ( π x ) π x × ∑ i ≥ 1 ζ ( 2 i ) x i . {\displaystyle {\frac {\pi ^{2k}}{2}}\cdot {\frac {(2k)\cdot (-1)^{k}}{(2k+1)!}}=-[x^{2k}]{\frac {\sin(\pi x)}{\pi x}}\times \sum _{i\geq 1}\zeta (2i)x^{i}.} </p><p>Then by differentiation and rearrangement of the terms in the previous equation, we obtain that ζ ( 2 k ) = [ x 2 k ] 1 2 ( 1 − π x cot ( π x ) ) . {\displaystyle \zeta (2k)=[x^{2k}]{\frac {1}{2}}\left(1-\pi x\cot(\pi x)\right).} </p> <h3>Consequences of Euler's proof</h3> <p>By the above results, we can conclude that ζ ( 2 k ) {\displaystyle \zeta (2k)} is <i>always</i> a <a href="/facts/Rational/ueS0w9lk">rational</a> multiple of π 2 k {\displaystyle \pi ^{2k}} . In particular, since π {\displaystyle \pi } and integer powers of it are <a href="/facts/Transcendental_number/IrB7JppM">transcendental</a>, we can conclude at this point that ζ ( 2 k ) {\displaystyle \zeta (2k)} is <a href="/facts/Irrational/UQEGMC85">irrational</a>, and more precisely, <a href="/facts/Transcendental_number/IrB7JppM">transcendental</a> for all k ≥ 1 {\displaystyle k\geq 1} . By contrast, the properties of the odd-indexed <a href="/facts/Zeta_constants/3W4ClRUP">zeta constants</a>, including <a href="/facts/Ap%25C3%25A9ry%2527s_constant/59do2TbQ">Apéry's constant</a> ζ ( 3 ) {\displaystyle \zeta (3)} , are almost completely unknown. </p> <h2 id="the-riemann-zeta-function">The Riemann zeta function</h2> <p>The <a href="/facts/Riemann_zeta_function/MnfRu7lY">Riemann zeta function</a> <i>ζ</i>(<i>s</i>) is one of the most significant functions in mathematics because of its relationship to the distribution of the <a href="/facts/Prime_number/L20FLkgn">prime numbers</a>. The zeta function is defined for any <a href="/facts/Complex_number/2w7mqI7e">complex number</a> <i>s</i> with real part greater than 1 by the following formula: ζ ( s ) = ∑ n = 1 ∞ 1 n s . {\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}.} </p><p>Taking <i>s</i> = 2, we see that <i>ζ</i>(2) is equal to the sum of the reciprocals of the squares of all positive integers: ζ ( 2 ) = ∑ n = 1 ∞ 1 n 2 = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 2 6 ≈ 1.644934. {\displaystyle \zeta (2)=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}\approx 1.644934.} </p><p>Convergence can be proven by the <a href="/facts/Integral_test_for_convergence/7KRhJwRr">integral test</a>, or by the following inequality: ∑ n = 1 N 1 n 2 < 1 + ∑ n = 2 N 1 n ( n − 1 ) = 1 + ∑ n = 2 N ( 1 n − 1 − 1 n ) = 1 + 1 − 1 N ⟶ N → ∞ 2. {\displaystyle {\begin{aligned}\sum _{n=1}^{N}{\frac {1}{n^{2}}}&<1+\sum _{n=2}^{N}{\frac {1}{n(n-1)}}\\&=1+\sum _{n=2}^{N}\left({\frac {1}{n-1}}-{\frac {1}{n}}\right)\\&=1+1-{\frac {1}{N}}\;{\stackrel {N\to \infty }{\longrightarrow }}\;2.\end{aligned}}} </p><p>This gives us the <a href="/facts/Upper_bound/YJje9oC2">upper bound</a> 2, and because the infinite sum contains no negative terms, it must converge to a value strictly between 0 and 2. It can be shown that <i>ζ</i>(<i>s</i>) has a simple expression in terms of the <a href="/facts/Bernoulli_number/RH0mblhs">Bernoulli numbers</a> whenever <i>s</i> is a positive even integer. With <i>s</i> = 2<i>n</i>:<a class="footnote-ref" id="fnref:9" href="#fn:9"><sup>9</sup></a> ζ ( 2 n ) = ( 2 π ) 2 n ( − 1 ) n + 1 B 2 n 2 ⋅ ( 2 n ) ! . {\displaystyle \zeta (2n)={\frac {(2\pi )^{2n}(-1)^{n+1}B_{2n}}{2\cdot (2n)!}}.} </p> <h2 id="a-proof-using-eulers-formula-and-lhpitals-rule">A proof using Euler's formula and L'Hôpital's rule</h2> <p>The normalized <a href="/facts/Sinc_function/lSvDtHy0">sinc function</a> sinc ( x ) = sin ( π x ) π x {\displaystyle {\text{sinc}}(x)={\frac {\sin(\pi x)}{\pi x}}} has a <a href="/facts/Weierstrass_factorization_theorem/atdInY5m">Weierstrass factorization</a> representation as an infinite product: sin ( π x ) π x = ∏ n = 1 ∞ ( 1 − x 2 n 2 ) . {\displaystyle {\frac {\sin(\pi x)}{\pi x}}=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}}}\right).} </p><p>The infinite product is <a href="/facts/Analytic_function/JhL3z8FP">analytic</a>, so taking the <a href="/facts/Natural_logarithm/PnreKEzI">natural logarithm</a> of both sides and differentiating yields π cos ( π x ) sin ( π x ) − 1 x = − ∑ n = 1 ∞ 2 x n 2 − x 2 {\displaystyle {\frac {\pi \cos(\pi x)}{\sin(\pi x)}}-{\frac {1}{x}}=-\sum _{n=1}^{\infty }{\frac {2x}{n^{2}-x^{2}}}} </p><p>(by <a href="/facts/Uniform_convergence/ecz9eNWn">uniform convergence</a>, the interchange of the derivative and infinite series is permissible). After dividing the equation by 2 x {\displaystyle 2x} and regrouping one gets 1 2 x 2 − π cot ( π x ) 2 x = ∑ n = 1 ∞ 1 n 2 − x 2 . {\displaystyle {\frac {1}{2x^{2}}}-{\frac {\pi \cot(\pi x)}{2x}}=\sum _{n=1}^{\infty }{\frac {1}{n^{2}-x^{2}}}.} </p><p>We make a change of variables ( x = − i t {\displaystyle x=-it} ): − 1 2 t 2 + π cot ( − π i t ) 2 i t = ∑ n = 1 ∞ 1 n 2 + t 2 . {\displaystyle -{\frac {1}{2t^{2}}}+{\frac {\pi \cot(-\pi it)}{2it}}=\sum _{n=1}^{\infty }{\frac {1}{n^{2}+t^{2}}}.} </p><p><a href="/facts/Euler%2527s_formula/Wq7VXIV0">Euler's formula</a> can be used to deduce that π cot ( − π i t ) 2 i t = π 2 i t i ( e 2 π t + 1 ) e 2 π t − 1 = π 2 t + π t ( e 2 π t − 1 ) . {\displaystyle {\frac {\pi \cot(-\pi it)}{2it}}={\frac {\pi }{2it}}{\frac {i\left(e^{2\pi t}+1\right)}{e^{2\pi t}-1}}={\frac {\pi }{2t}}+{\frac {\pi }{t\left(e^{2\pi t}-1\right)}}.} or using the corresponding <a href="/facts/Hyperbolic_function/Y4Cb5S35">hyperbolic function</a>: π cot ( − π i t ) 2 i t = π 2 t i cot ( π i t ) = π 2 t coth ( π t ) . {\displaystyle {\frac {\pi \cot(-\pi it)}{2it}}={\frac {\pi }{2t}}{i\cot(\pi it)}={\frac {\pi }{2t}}\coth(\pi t).} </p><p>Then ∑ n = 1 ∞ 1 n 2 + t 2 = π ( t e 2 π t + t ) − e 2 π t + 1 2 ( t 2 e 2 π t − t 2 ) = − 1 2 t 2 + π 2 t coth ( π t ) . {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}+t^{2}}}={\frac {\pi \left(te^{2\pi t}+t\right)-e^{2\pi t}+1}{2\left(t^{2}e^{2\pi t}-t^{2}\right)}}=-{\frac {1}{2t^{2}}}+{\frac {\pi }{2t}}\coth(\pi t).} </p><p>Now we take the <a href="/facts/Limit_(mathematics)/j8JBhj4e">limit</a> as t {\displaystyle t} approaches zero and use <a href="/facts/L%2527H%25C3%25B4pital%2527s_rule/pCXK75Io">L'Hôpital's rule</a> thrice. By <a href="/facts/Tannery%2527s_theorem/YGXRLIy7">Tannery's theorem</a> applied to lim t → ∞ ∑ n = 1 ∞ 1 / ( n 2 + 1 / t 2 ) {\textstyle \lim _{t\to \infty }\sum _{n=1}^{\infty }1/(n^{2}+1/t^{2})} , we can <a href="/facts/Interchange_of_limiting_operations/qqW0RnEn">interchange the limit and infinite series</a> so that lim t → 0 ∑ n = 1 ∞ 1 / ( n 2 + t 2 ) = ∑ n = 1 ∞ 1 / n 2 {\textstyle \lim _{t\to 0}\sum _{n=1}^{\infty }1/(n^{2}+t^{2})=\sum _{n=1}^{\infty }1/n^{2}} and by L'Hôpital's rule ∑ n = 1 ∞ 1 n 2 = lim t → 0 π 4 2 π t e 2 π t − e 2 π t + 1 π t 2 e 2 π t + t e 2 π t − t = lim t → 0 π 3 t e 2 π t 2 π ( π t 2 e 2 π t + 2 t e 2 π t ) + e 2 π t − 1 = lim t → 0 π 2 ( 2 π t + 1 ) 4 π 2 t 2 + 12 π t + 6 = π 2 6 . {\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}&=\lim _{t\to 0}{\frac {\pi }{4}}{\frac {2\pi te^{2\pi t}-e^{2\pi t}+1}{\pi t^{2}e^{2\pi t}+te^{2\pi t}-t}}\\[6pt]&=\lim _{t\to 0}{\frac {\pi ^{3}te^{2\pi t}}{2\pi \left(\pi t^{2}e^{2\pi t}+2te^{2\pi t}\right)+e^{2\pi t}-1}}\\[6pt]&=\lim _{t\to 0}{\frac {\pi ^{2}(2\pi t+1)}{4\pi ^{2}t^{2}+12\pi t+6}}\\[6pt]&={\frac {\pi ^{2}}{6}}.\end{aligned}}} </p> <h2 id="a-proof-using-fourier-series">A proof using Fourier series</h2> <p>Use <a href="/facts/Parseval%2527s_identity/EukqvgUp">Parseval's identity</a> (applied to the function <i>f</i>(<i>x</i>) = <i>x</i>) to obtain ∑ n = − ∞ ∞ | c n | 2 = 1 2 π ∫ − π π x 2 d x , {\displaystyle \sum _{n=-\infty }^{\infty }|c_{n}|^{2}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }x^{2}\,dx,} where c n = 1 2 π ∫ − π π x e − i n x d x = n π cos ( n π ) − sin ( n π ) π n 2 i = cos ( n π ) n i = ( − 1 ) n n i {\displaystyle {\begin{aligned}c_{n}&={\frac {1}{2\pi }}\int _{-\pi }^{\pi }xe^{-inx}\,dx\\[4pt]&={\frac {n\pi \cos(n\pi )-\sin(n\pi )}{\pi n^{2}}}i\\[4pt]&={\frac {\cos(n\pi )}{n}}i\\[4pt]&={\frac {(-1)^{n}}{n}}i\end{aligned}}} </p><p>for <i>n</i> ≠ 0, and <i>c</i>0 = 0. Thus, | c n | 2 = { 1 n 2 , for n ≠ 0 , 0 , for n = 0 , {\displaystyle |c_{n}|^{2}={\begin{cases}{\dfrac {1}{n^{2}}},&{\text{for }}n\neq 0,\\0,&{\text{for }}n=0,\end{cases}}} </p><p>and ∑ n = − ∞ ∞ | c n | 2 = 2 ∑ n = 1 ∞ 1 n 2 = 1 2 π ∫ − π π x 2 d x . {\displaystyle \sum _{n=-\infty }^{\infty }|c_{n}|^{2}=2\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }x^{2}\,dx.} </p><p>Therefore, ∑ n = 1 ∞ 1 n 2 = 1 4 π ∫ − π π x 2 d x = π 2 6 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{4\pi }}\int _{-\pi }^{\pi }x^{2}\,dx={\frac {\pi ^{2}}{6}}} as required. </p> <h2 id="another-proof-using-parsevals-identity">Another proof using Parseval's identity</h2> <p>Given a <a href="/facts/Complete_orthonormal_basis/pzfKHlJG">complete orthonormal basis</a> in the space L per 2 ( 0 , 1 ) {\displaystyle L_{\operatorname {per} }^{2}(0,1)} of <a href="/facts/Lp_space/gbad0Rv1">L2</a> <a href="/facts/Periodic_function/Sm8lJQsF">periodic functions</a> over ( 0 , 1 ) {\displaystyle (0,1)} (i.e., the subspace of <a href="/facts/Square-integrable_function/IASblcJJ">square-integrable functions</a> which are also <a href="/facts/Periodic_function/Sm8lJQsF">periodic</a>), denoted by { e i } i = − ∞ ∞ {\displaystyle \{e_{i}\}_{i=-\infty }^{\infty }} , <a href="/facts/Parseval%2527s_identity/EukqvgUp">Parseval's identity</a> tells us that ‖ x ‖ 2 = ∑ i = − ∞ ∞ | ⟨ e i , x ⟩ | 2 , {\displaystyle \|x\|^{2}=\sum _{i=-\infty }^{\infty }|\langle e_{i},x\rangle |^{2},} </p><p>where ‖ x ‖ := ⟨ x , x ⟩ {\displaystyle \|x\|:={\sqrt {\langle x,x\rangle }}} is defined in terms of the <a href="/facts/Inner_product/HoyjElEy">inner product</a> on this <a href="/facts/Hilbert_space/aDFMh4lV">Hilbert space</a> given by ⟨ f , g ⟩ = ∫ 0 1 f ( x ) g ( x ) ¯ d x , f , g ∈ L per 2 ( 0 , 1 ) . {\displaystyle \langle f,g\rangle =\int _{0}^{1}f(x){\overline {g(x)}}\,dx,\ f,g\in L_{\operatorname {per} }^{2}(0,1).} </p><p>We can consider the <a href="/facts/Orthonormal_basis/pzfKHlJG">orthonormal basis</a> on this space defined by e k ≡ e k ( ϑ ) := exp ( 2 π ı k ϑ ) {\displaystyle e_{k}\equiv e_{k}(\vartheta ):=\exp(2\pi \imath k\vartheta )} such that ⟨ e k , e j ⟩ = ∫ 0 1 e 2 π ı ( k − j ) ϑ d ϑ = δ k , j {\displaystyle \langle e_{k},e_{j}\rangle =\int _{0}^{1}e^{2\pi \imath (k-j)\vartheta }\,d\vartheta =\delta _{k,j}} . Then if we take f ( ϑ ) := ϑ {\displaystyle f(\vartheta ):=\vartheta } , we can compute both that ‖ f ‖ 2 = ∫ 0 1 ϑ 2 d ϑ = 1 3 ⟨ f , e k ⟩ = ∫ 0 1 ϑ e − 2 π ı k ϑ d ϑ = { 1 2 , k = 0 − 1 2 π ı k k ≠ 0 , {\displaystyle {\begin{aligned}\|f\|^{2}&=\int _{0}^{1}\vartheta ^{2}\,d\vartheta ={\frac {1}{3}}\\\langle f,e_{k}\rangle &=\int _{0}^{1}\vartheta e^{-2\pi \imath k\vartheta }\,d\vartheta ={\Biggl \{}{\begin{array}{ll}{\frac {1}{2}},&k=0\\-{\frac {1}{2\pi \imath k}}&k\neq 0,\end{array}}\end{aligned}}} </p><p>by elementary <a href="/facts/Calculus/MEmUTYoz">calculus</a> and <a href="/facts/Integration_by_parts/aV6JgBNd">integration by parts</a>, respectively. Finally, by <a href="/facts/Parseval%2527s_identity/EukqvgUp">Parseval's identity</a> stated in the form above, we obtain that ‖ f ‖ 2 = 1 3 = ∑ k ≠ 0 k = − ∞ ∞ 1 ( 2 π k ) 2 + 1 4 = 2 ∑ k = 1 ∞ 1 ( 2 π k ) 2 + 1 4 ⟹ π 2 6 = 2 π 2 3 − π 2 2 = ζ ( 2 ) . {\displaystyle {\begin{aligned}\|f\|^{2}={\frac {1}{3}}&=\sum _{\stackrel {k=-\infty }{k\neq 0}}^{\infty }{\frac {1}{(2\pi k)^{2}}}+{\frac {1}{4}}=2\sum _{k=1}^{\infty }{\frac {1}{(2\pi k)^{2}}}+{\frac {1}{4}}\\&\implies {\frac {\pi ^{2}}{6}}={\frac {2\pi ^{2}}{3}}-{\frac {\pi ^{2}}{2}}=\zeta (2).\end{aligned}}} </p> <h3>Generalizations and recurrence relations</h3> <p>Note that by considering higher-order powers of f j ( ϑ ) := ϑ j ∈ L per 2 ( 0 , 1 ) {\displaystyle f_{j}(\vartheta ):=\vartheta ^{j}\in L_{\operatorname {per} }^{2}(0,1)} we can use <a href="/facts/Integration_by_parts/aV6JgBNd">integration by parts</a> to extend this method to enumerating formulas for ζ ( 2 j ) {\displaystyle \zeta (2j)} when j > 1 {\displaystyle j>1} . In particular, suppose we let I j , k := ∫ 0 1 ϑ j e − 2 π ı k ϑ d ϑ , {\displaystyle I_{j,k}:=\int _{0}^{1}\vartheta ^{j}e^{-2\pi \imath k\vartheta }\,d\vartheta ,} </p><p>so that <a href="/facts/Integration_by_parts/aV6JgBNd">integration by parts</a> yields the <a href="/facts/Recurrence_relation/JolXC71M">recurrence relation</a> that I j , k = { 1 j + 1 , k = 0 ; − 1 2 π ı ⋅ k + j 2 π ı ⋅ k I j − 1 , k , k ≠ 0 = { 1 j + 1 , k = 0 ; − ∑ m = 1 j j ! ( j + 1 − m ) ! ⋅ 1 ( 2 π ı ⋅ k ) m , k ≠ 0. {\displaystyle {\begin{aligned}I_{j,k}&={\begin{cases}{\frac {1}{j+1}},&k=0;\\[4pt]-{\frac {1}{2\pi \imath \cdot k}}+{\frac {j}{2\pi \imath \cdot k}}I_{j-1,k},&k\neq 0\end{cases}}\\[6pt]&={\begin{cases}{\frac {1}{j+1}},&k=0;\\[4pt]-\sum \limits _{m=1}^{j}{\frac {j!}{(j+1-m)!}}\cdot {\frac {1}{(2\pi \imath \cdot k)^{m}}},&k\neq 0.\end{cases}}\end{aligned}}} </p><p>Then by applying <a href="/facts/Parseval%2527s_identity/EukqvgUp">Parseval's identity</a> as we did for the first case above along with the linearity of the <a href="/facts/Inner_product/HoyjElEy">inner product</a> yields that ‖ f j ‖ 2 = 1 2 j + 1 = 2 ∑ k ≥ 1 I j , k I ¯ j , k + 1 ( j + 1 ) 2 = 2 ∑ m = 1 j ∑ r = 1 j j ! 2 ( j + 1 − m ) ! ( j + 1 − r ) ! ( − 1 ) r ı m + r ζ ( m + r ) ( 2 π ) m + r + 1 ( j + 1 ) 2 . {\displaystyle {\begin{aligned}\|f_{j}\|^{2}={\frac {1}{2j+1}}&=2\sum _{k\geq 1}I_{j,k}{\bar {I}}_{j,k}+{\frac {1}{(j+1)^{2}}}\\[6pt]&=2\sum _{m=1}^{j}\sum _{r=1}^{j}{\frac {j!^{2}}{(j+1-m)!(j+1-r)!}}{\frac {(-1)^{r}}{\imath ^{m+r}}}{\frac {\zeta (m+r)}{(2\pi )^{m+r}}}+{\frac {1}{(j+1)^{2}}}.\end{aligned}}} </p> <h2 id="proof-using-differentiation-under-the-integral-sign">Proof using differentiation under the integral sign</h2> <p>It's possible to prove the result using elementary calculus by applying the <a href="/facts/Leibniz_integral_rule/bFKo8h0I">differentiation under the integral sign</a> technique to an integral due to Freitas:<a class="footnote-ref" id="fnref:10" href="#fn:10"><sup>10</sup></a> I ( α ) = ∫ 0 ∞ ln ( 1 + α e − x + e − 2 x ) d x . {\displaystyle I(\alpha )=\int _{0}^{\infty }\ln \left(1+\alpha e^{-x}+e^{-2x}\right)dx.} </p><p>While the <a href="/facts/Primitive_function/5qDsudIv">primitive function</a> of the integrand cannot be expressed in terms of elementary functions, by differentiating with respect to α {\displaystyle \alpha } we arrive at </p><p> d I d α = ∫ 0 ∞ e − x 1 + α e − x + e − 2 x d x , {\displaystyle {\frac {dI}{d\alpha }}=\int _{0}^{\infty }{\frac {e^{-x}}{1+\alpha e^{-x}+e^{-2x}}}dx,} which can be integrated by <a href="/facts/Integration_by_substitution/NvXfTTaf">substituting</a> u = e − x {\displaystyle u=e^{-x}} and decomposing into <a href="/facts/Partial_fractions/KLRmM3Hw">partial fractions</a>. In the range − 2 ≤ α ≤ 2 {\displaystyle -2\leq \alpha \leq 2} the definite integral reduces to </p><p> d I d α = 2 4 − α 2 [ arctan ( α + 2 4 − α 2 ) − arctan ( α 4 − α 2 ) ] . {\displaystyle {\frac {dI}{d\alpha }}={\frac {2}{\sqrt {4-\alpha ^{2}}}}\left[\arctan \left({\frac {\alpha +2}{\sqrt {4-\alpha ^{2}}}}\right)-\arctan \left({\frac {\alpha }{\sqrt {4-\alpha ^{2}}}}\right)\right].} </p><p>The expression can be simplified using the <a href="/facts/Inverse_trigonometric_functions/bVwU9FPG">arctangent addition formula</a> and integrated with respect to α {\displaystyle \alpha } by means of <a href="/facts/Trigonometric_substitution/U6LBLAYU">trigonometric substitution</a>, resulting in </p><p> I ( α ) = − 1 2 arccos ( α 2 ) 2 + c . {\displaystyle I(\alpha )=-{\frac {1}{2}}\arccos \left({\frac {\alpha }{2}}\right)^{2}+c.} </p><p>The <a href="/facts/Integration_constant/dFzDzFhl">integration constant</a> c {\displaystyle c} can be determined by noticing that two distinct values of I ( α ) {\displaystyle I(\alpha )} are related by </p><p> I ( 2 ) = 4 I ( 0 ) , {\displaystyle I(2)=4I(0),} because when calculating I ( 2 ) {\displaystyle I(2)} we can <a href="/facts/Factorization/tXrBXFh1">factor</a> 1 + 2 e − x + e − 2 x = ( 1 + e − x ) 2 {\displaystyle 1+2e^{-x}+e^{-2x}=(1+e^{-x})^{2}} and express it in terms of I ( 0 ) {\displaystyle I(0)} using the <a href="/facts/List_of_logarithmic_identities/5T8DX7d8">logarithm of a power identity</a> and the <a href="/facts/Integration_by_substitution/NvXfTTaf">substitution</a> u = x / 2 {\displaystyle u=x/2} . This makes it possible to determine c = π 2 6 {\displaystyle c={\frac {\pi ^{2}}{6}}} , and it follows that </p><p> I ( − 2 ) = 2 ∫ 0 ∞ ln ( 1 − e − x ) d x = − π 2 3 . {\displaystyle I(-2)=2\int _{0}^{\infty }\ln(1-e^{-x})dx=-{\frac {\pi ^{2}}{3}}.} </p><p>This final integral can be evaluated by expanding the natural logarithm into its <a href="/facts/Mercator_series/BYxgSMVA">Taylor series</a>: </p><p> ∫ 0 ∞ ln ( 1 − e − x ) d x = − ∑ n = 1 ∞ ∫ 0 ∞ e − n x n d x = − ∑ n = 1 ∞ 1 n 2 . {\displaystyle \int _{0}^{\infty }\ln(1-e^{-x})dx=-\sum _{n=1}^{\infty }\int _{0}^{\infty }{\frac {e^{-nx}}{n}}dx=-\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}.} </p><p>The last two identities imply </p><p> ∑ n = 1 ∞ 1 n 2 = π 2 6 . {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}.} </p> <h2 id="cauchys-proof">Cauchy's proof</h2> <p>While most proofs use results from advanced <a href="/facts/Mathematics/pxTouaz4">mathematics</a>, such as <a href="/facts/Fourier_analysis/sUA8JP6q">Fourier analysis</a>, <a href="/facts/Complex_analysis/hgCoQlH0">complex analysis</a>, and <a href="/facts/Multivariable_calculus/PEs5SgUF">multivariable calculus</a>, the following does not even require single-variable <a href="/facts/Calculus/MEmUTYoz">calculus</a> (until a single <a href="/facts/Limit_of_a_function/HFbQ69e5">limit</a> is taken at the end). </p><p>For a proof using the residue theorem, see <a href="/facts/Residue_theorem/M5ITFIyx">here</a>. </p> <h3>History of this proof</h3> <p>The proof goes back to <a href="/facts/Augustin_Louis_Cauchy/aoTjgj6j">Augustin Louis Cauchy</a> (Cours d'Analyse, 1821, Note VIII). In 1954, this proof appeared in the book of <a href="/facts/Akiva_Yaglom/drLF50I3">Akiva</a> and <a href="/facts/Isaak_Yaglom/5f6jeMN4">Isaak Yaglom</a> "Nonelementary Problems in an Elementary Exposition". Later, in 1982, it appeared in the journal <i>Eureka</i>,<a class="footnote-ref" id="fnref:11" href="#fn:11"><sup>11</sup></a> attributed to John Scholes, but Scholes claims he learned the proof from <a href="/facts/Peter_Swinnerton-Dyer/D66SILlt">Peter Swinnerton-Dyer</a>, and in any case he maintains the proof was "common knowledge at <a href="/facts/University_of_Cambridge/0la7aX9o">Cambridge</a> in the late 1960s".<a class="footnote-ref" id="fnref:12" href="#fn:12"><sup>12</sup></a> </p> <h3>The proof</h3> <p>The main idea behind the proof is to bound the partial (finite) sums ∑ k = 1 m 1 k 2 = 1 1 2 + 1 2 2 + ⋯ + 1 m 2 {\displaystyle \sum _{k=1}^{m}{\frac {1}{k^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{m^{2}}}} between two expressions, each of which will tend to π2/6 as <i>m</i> approaches infinity. The two expressions are derived from identities involving the <a href="/facts/Cotangent/czej7ur0">cotangent</a> and <a href="/facts/Cosecant/czej7ur0">cosecant</a> functions. These identities are in turn derived from <a href="/facts/De_Moivre%2527s_formula/uyG4k2uC">de Moivre's formula</a>, and we now turn to establishing these identities. </p><p>Let <i>x</i> be a real number with 0 < <i>x</i> < π/2, and let <i>n</i> be a positive odd integer. Then from de Moivre's formula and the definition of the cotangent function, we have cos ( n x ) + i sin ( n x ) sin n x = ( cos x + i sin x ) n sin n x = ( cos x + i sin x sin x ) n = ( cot x + i ) n . {\displaystyle {\begin{aligned}{\frac {\cos(nx)+i\sin(nx)}{\sin ^{n}x}}&={\frac {(\cos x+i\sin x)^{n}}{\sin ^{n}x}}\\[4pt]&=\left({\frac {\cos x+i\sin x}{\sin x}}\right)^{n}\\[4pt]&=(\cot x+i)^{n}.\end{aligned}}} </p><p>From the <a href="/facts/Binomial_theorem/hd4SmkUd">binomial theorem</a>, we have ( cot x + i ) n = ( n 0 ) cot n x + ( n 1 ) ( cot n − 1 x ) i + ⋯ + ( n n − 1 ) ( cot x ) i n − 1 + ( n n ) i n = ( ( n 0 ) cot n x − ( n 2 ) cot n − 2 x ± ⋯ ) + i ( ( n 1 ) cot n − 1 x − ( n 3 ) cot n − 3 x ± ⋯ ) . {\displaystyle {\begin{aligned}(\cot x+i)^{n}=&{n \choose 0}\cot ^{n}x+{n \choose 1}(\cot ^{n-1}x)i+\cdots +{n \choose {n-1}}(\cot x)i^{n-1}+{n \choose n}i^{n}\\[6pt]=&{\Bigg (}{n \choose 0}\cot ^{n}x-{n \choose 2}\cot ^{n-2}x\pm \cdots {\Bigg )}\;+\;i{\Bigg (}{n \choose 1}\cot ^{n-1}x-{n \choose 3}\cot ^{n-3}x\pm \cdots {\Bigg )}.\end{aligned}}} </p><p>Combining the two equations and equating imaginary parts gives the identity sin ( n x ) sin n x = ( ( n 1 ) cot n − 1 x − ( n 3 ) cot n − 3 x ± ⋯ ) . {\displaystyle {\frac {\sin(nx)}{\sin ^{n}x}}={\Bigg (}{n \choose 1}\cot ^{n-1}x-{n \choose 3}\cot ^{n-3}x\pm \cdots {\Bigg )}.} </p><p>We take this identity, fix a positive integer <i>m</i>, set <i>n</i> = 2<i>m</i> + 1, and consider <i>xr</i> = <i>r</i>π/2<i>m</i> + 1 for <i>r</i> = 1, 2, ..., <i>m</i>. Then <i>nxr</i> is a multiple of π and therefore sin(<i>nxr</i>) = 0. So, 0 = ( 2 m + 1 1 ) cot 2 m x r − ( 2 m + 1 3 ) cot 2 m − 2 x r ± ⋯ + ( − 1 ) m ( 2 m + 1 2 m + 1 ) {\displaystyle 0={{2m+1} \choose 1}\cot ^{2m}x_{r}-{{2m+1} \choose 3}\cot ^{2m-2}x_{r}\pm \cdots +(-1)^{m}{{2m+1} \choose {2m+1}}} </p><p>for every <i>r</i> = 1, 2, ..., <i>m</i>. The values <i>xr</i> = <i>x</i>1, <i>x</i>2, ..., <i>xm</i> are distinct numbers in the interval 0 < <i>xr</i> < π/2. Since the function cot2 <i>x</i> is <a href="/facts/Injective_function/5ES6tqf5">one-to-one</a> on this interval, the numbers <i>tr</i> = cot2 <i>xr</i> are distinct for <i>r</i> = 1, 2, ..., <i>m</i>. By the above equation, these <i>m</i> numbers are the roots of the <i>m</i>th degree polynomial p ( t ) = ( 2 m + 1 1 ) t m − ( 2 m + 1 3 ) t m − 1 ± ⋯ + ( − 1 ) m ( 2 m + 1 2 m + 1 ) . {\displaystyle p(t)={{2m+1} \choose 1}t^{m}-{{2m+1} \choose 3}t^{m-1}\pm \cdots +(-1)^{m}{{2m+1} \choose {2m+1}}.} </p><p>By <a href="/facts/Vieta%2527s_formulas/u3eQJWkA">Vieta's formulas</a> we can calculate the sum of the roots directly by examining the first two coefficients of the polynomial, and this comparison shows that cot 2 x 1 + cot 2 x 2 + ⋯ + cot 2 x m = ( 2 m + 1 3 ) ( 2 m + 1 1 ) = 2 m ( 2 m − 1 ) 6 . {\displaystyle \cot ^{2}x_{1}+\cot ^{2}x_{2}+\cdots +\cot ^{2}x_{m}={\frac {\binom {2m+1}{3}}{\binom {2m+1}{1}}}={\frac {2m(2m-1)}{6}}.} </p><p>Substituting the <a href="/facts/List_of_trigonometric_identities/Xw9Jx8Vz">identity</a> csc2 <i>x</i> = cot2 <i>x</i> + 1, we have csc 2 x 1 + csc 2 x 2 + ⋯ + csc 2 x m = 2 m ( 2 m − 1 ) 6 + m = 2 m ( 2 m + 2 ) 6 . {\displaystyle \csc ^{2}x_{1}+\csc ^{2}x_{2}+\cdots +\csc ^{2}x_{m}={\frac {2m(2m-1)}{6}}+m={\frac {2m(2m+2)}{6}}.} </p><p>Now consider the inequality cot2 <i>x</i> < 1/<i>x</i>2 < csc2 <i>x</i> (illustrated geometrically above). If we add up all these inequalities for each of the numbers <i>xr</i> = <i>r</i>π/2<i>m</i> + 1, and if we use the two identities above, we get 2 m ( 2 m − 1 ) 6 < ( 2 m + 1 π ) 2 + ( 2 m + 1 2 π ) 2 + ⋯ + ( 2 m + 1 m π ) 2 < 2 m ( 2 m + 2 ) 6 . {\displaystyle {\frac {2m(2m-1)}{6}}<\left({\frac {2m+1}{\pi }}\right)^{2}+\left({\frac {2m+1}{2\pi }}\right)^{2}+\cdots +\left({\frac {2m+1}{m\pi }}\right)^{2}<{\frac {2m(2m+2)}{6}}.} </p><p>Multiplying through by (π/2<i>m</i> + 1)2, this becomes π 2 6 ( 2 m 2 m + 1 ) ( 2 m − 1 2 m + 1 ) < 1 1 2 + 1 2 2 + ⋯ + 1 m 2 < π 2 6 ( 2 m 2 m + 1 ) ( 2 m + 2 2 m + 1 ) . {\displaystyle {\frac {\pi ^{2}}{6}}\left({\frac {2m}{2m+1}}\right)\left({\frac {2m-1}{2m+1}}\right)<{\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{m^{2}}}<{\frac {\pi ^{2}}{6}}\left({\frac {2m}{2m+1}}\right)\left({\frac {2m+2}{2m+1}}\right).} </p><p>As <i>m</i> approaches infinity, the left and right hand expressions each approach π2/6, so by the <a href="/facts/Squeeze_theorem/yo0YSBkK">squeeze theorem</a>, ζ ( 2 ) = ∑ k = 1 ∞ 1 k 2 = lim m → ∞ ( 1 1 2 + 1 2 2 + ⋯ + 1 m 2 ) = π 2 6 {\displaystyle \zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}=\lim _{m\to \infty }\left({\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{m^{2}}}\right)={\frac {\pi ^{2}}{6}}} </p><p>and this completes the proof. </p> <h2 id="proof-assuming-weils-conjecture-on-tamagawa-numbers">Proof assuming Weil's conjecture on Tamagawa numbers</h2> <p>A proof is also possible assuming <a href="/facts/Weil%2527s_conjecture_on_Tamagawa_numbers/XP1Jeqf4">Weil's conjecture on Tamagawa numbers</a>.<a class="footnote-ref" id="fnref:13" href="#fn:13"><sup>13</sup></a> The conjecture asserts for the case of the <a href="/facts/Algebraic_group/DDtBCIvh">algebraic group</a> SL2(R) that the <a href="/facts/Tamagawa_number/NRYVaI7S">Tamagawa number</a> of the group is one. That is, the quotient of the special linear group over the rational <a href="/facts/Adele_ring/2LsbgeBE">adeles</a> by the special linear group of the rationals (a <a href="/facts/Compact_set/d0cgXJH7">compact set</a>, because S L 2 ( Q ) {\displaystyle SL_{2}(\mathbb {Q} )} is a lattice in the adeles) has Tamagawa measure 1: τ ( S L 2 ( Q ) ∖ S L 2 ( A Q ) ) = 1. {\displaystyle \tau (SL_{2}(\mathbb {Q} )\setminus SL_{2}(A_{\mathbb {Q} }))=1.} </p><p>To determine a Tamagawa measure, the group S L 2 {\displaystyle SL_{2}} consists of matrices [ x y z t ] {\displaystyle {\begin{bmatrix}x&y\\z&t\end{bmatrix}}} with x t − y z = 1 {\displaystyle xt-yz=1} . An invariant <a href="/facts/Volume_form/2x37pDLg">volume form</a> on the group is ω = 1 x d x ∧ d y ∧ d z . {\displaystyle \omega ={\frac {1}{x}}dx\wedge dy\wedge dz.} </p><p>The measure of the quotient is the product of the measures of S L 2 ( Z ) ∖ S L 2 ( R ) {\displaystyle SL_{2}(\mathbb {Z} )\setminus SL_{2}(\mathbb {R} )} corresponding to the infinite place, and the measures of S L 2 ( Z p ) {\displaystyle SL_{2}(\mathbb {Z} _{p})} in each finite place, where Z p {\displaystyle \mathbb {Z} _{p}} is the <a href="/facts/P-adic_integers/SDgDbkLF">p-adic integers</a>. </p><p>For the local factors, ω ( S L 2 ( Z p ) ) = | S L 2 ( F p ) | ω ( S L 2 ( Z p , p ) ) {\displaystyle \omega (SL_{2}(\mathbb {Z} _{p}))=|SL_{2}(F_{p})|\omega (SL_{2}(\mathbb {Z} _{p},p))} where F p {\displaystyle F_{p}} is the field with p {\displaystyle p} elements, and S L 2 ( Z p , p ) {\displaystyle SL_{2}(\mathbb {Z} _{p},p)} is the <a href="/facts/Congruence_subgroup/V334rHwz">congruence subgroup</a> modulo p {\displaystyle p} . Since each of the coordinates x , y , z {\displaystyle x,y,z} map the latter group onto p Z p {\displaystyle p\mathbb {Z} _{p}} and | 1 x | p = 1 {\displaystyle \left|{\frac {1}{x}}\right|_{p}=1} , the measure of S L 2 ( Z p , p ) {\displaystyle SL_{2}(\mathbb {Z} _{p},p)} is μ p ( p Z p ) 3 = p − 3 {\displaystyle \mu _{p}(p\mathbb {Z} _{p})^{3}=p^{-3}} , where μ p {\displaystyle \mu _{p}} is the normalized <a href="/facts/Haar_measure/2IcmTcuU">Haar measure</a> on Z p {\displaystyle \mathbb {Z} _{p}} . Also, a standard computation shows that | S L 2 ( F p ) | = p ( p 2 − 1 ) {\displaystyle |SL_{2}(F_{p})|=p(p^{2}-1)} . Putting these together gives ω ( S L 2 ( Z p ) ) = ( 1 − 1 / p 2 ) {\displaystyle \omega (SL_{2}(\mathbb {Z} _{p}))=(1-1/p^{2})} . </p><p>At the infinite place, an integral computation over the fundamental domain of S L 2 ( Z ) {\displaystyle SL_{2}(\mathbb {Z} )} shows that ω ( S L 2 ( Z ) ∖ S L 2 ( R ) = π 2 / 6 {\displaystyle \omega (SL_{2}(\mathbb {Z} )\setminus SL_{2}(\mathbb {R} )=\pi ^{2}/6} , and therefore the Weil conjecture finally gives 1 = π 2 6 ∏ p ( 1 − 1 p 2 ) . {\displaystyle 1={\frac {\pi ^{2}}{6}}\prod _{p}\left(1-{\frac {1}{p^{2}}}\right).} On the right-hand side, we recognize the <a href="/facts/Euler_product/Lvsv4U3M">Euler product</a> for 1 / ζ ( 2 ) {\displaystyle 1/\zeta (2)} , and so this gives the solution to the Basel problem. </p><p>This approach shows the connection between (hyperbolic) geometry and arithmetic, and can be inverted to give a proof of the Weil conjecture for the special case of S L 2 {\displaystyle SL_{2}} , contingent on an independent proof that ζ ( 2 ) = π 2 / 6 {\displaystyle \zeta (2)=\pi ^{2}/6} . </p> <h2 id="geometric-proof">Geometric proof</h2> <p>The Basel problem can be proved with <a href="/facts/Euclidean_geometry/V6cTcoCy">Euclidean geometry</a>, using the insight that <i>the real line can be seen as a circle of infinite radius</i>. An intuitive, if not completely rigorous, sketch is given here. </p> <ul><li>Choose an integer N {\displaystyle N} , and take N {\displaystyle N} equally spaced points on a circle with <i>circumference</i> equal to 2 N {\displaystyle 2N} . The radius of the circle is N / π {\displaystyle N/\pi } and the length of each <a href="/facts/Circular_arc/RrehhCQR">arc</a> between two points is 2 {\displaystyle 2} . Call the points P 1.. N {\displaystyle P_{1..N}} .</li> <li>Take another generic point Q {\displaystyle Q} on the circle, which will lie at a fraction 0 < α < 1 {\displaystyle 0<\alpha <1} of the arc between two consecutive points (say P 1 {\displaystyle P_{1}} and P 2 {\displaystyle P_{2}} without loss of generality).</li> <li>Draw all the <a href="/facts/Chord_(geometry)/aAyoEdLd">chords</a> joining Q {\displaystyle Q} with each of the P 1.. N {\displaystyle P_{1..N}} points. Now (this is the key to the proof), compute the <i>sum of the inverse squares</i> of the lengths of all these chords, call it s i s c {\displaystyle sisc} .</li> <li>The proof relies on the notable fact that (for a fixed α {\displaystyle \alpha } ), <i>the s i s c {\displaystyle sisc} does not depend on N {\displaystyle N} .</i> Note that intuitively, as N {\displaystyle N} increases, the number of chords increases, but their length increases too (as the circle gets bigger), so their inverse square decreases.</li> <li>In particular, take the case where α = 1 / 2 {\displaystyle \alpha =1/2} , meaning that Q {\displaystyle Q} is the midpoint of the arc between two consecutive P {\displaystyle P} 's. The s i s c {\displaystyle sisc} can then be found trivially from the case N = 1 {\displaystyle N=1} , where there is only one P {\displaystyle P} , and one Q {\displaystyle Q} on the opposite side of the circle. Then the chord is the diameter of the circle, of length 2 / π {\displaystyle 2/\pi } . The s i s c {\displaystyle sisc} is then π 2 / 4 {\displaystyle \pi ^{2}/4} .</li> <li>When N {\displaystyle N} goes to infinity, the circle approaches the real line. If you set the origin at Q {\displaystyle Q} , the points P 1.. N {\displaystyle P_{1..N}} are positioned at the <i>odd</i> integer positions (positive and negative), since the arcs have length 1 from Q {\displaystyle Q} to P 1 {\displaystyle P_{1}} , and 2 onward. You hence get this variation of the Basel Problem:</li></ul> <p> ∑ z = − ∞ ∞ 1 ( 2 z − 1 ) 2 = π 2 4 {\displaystyle \sum _{z=-\infty }^{\infty }{\frac {1}{(2z-1)^{2}}}={\frac {\pi ^{2}}{4}}} </p> <ul><li>From here, you can recover the original formulation with a bit of algebra, as:</li></ul> <p> ∑ n = 1 ∞ 1 n 2 = ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 + ∑ n = 1 ∞ 1 ( 2 n ) 2 = 1 2 ∑ z = − ∞ ∞ 1 ( 2 z − 1 ) 2 + 1 4 ∑ n = 1 ∞ 1 n 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}=\sum _{n=1}^{\infty }{\frac {1}{(2n-1)^{2}}}+\sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2}}\sum _{z=-\infty }^{\infty }{\frac {1}{(2z-1)^{2}}}+{\frac {1}{4}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}} </p><p>that is, </p><p> 3 4 ∑ n = 1 ∞ 1 n 2 = π 2 8 {\displaystyle {\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{8}}} </p><p>or </p><p> ∑ n = 1 ∞ 1 n 2 = π 2 6 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}} . </p><p>The independence of the s i s c {\displaystyle sisc} from N {\displaystyle N} can be proved easily with Euclidean geometry for the more restrictive case where N {\displaystyle N} is a power of 2, i.e. N = 2 n {\displaystyle N=2^{n}} , which still allows the limiting argument to be applied. The proof proceeds by <a href="/facts/Mathematical_induction/KxAPqNrH">induction</a> on n {\displaystyle n} , and uses the <a href="/facts/Inverse_Pythagorean_theorem/cbW99nSp">Inverse Pythagorean Theorem</a>, which states that: </p><p> 1 a 2 + 1 b 2 = 1 h 2 {\displaystyle {\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}={\frac {1}{h^{2}}}} </p><p>where a {\displaystyle a} and b {\displaystyle b} are the legs and h {\displaystyle h} is the height of a right triangle. </p> <ul><li>In the base case of n = 0 {\displaystyle n=0} , there is only 1 chord. In the case of α = 1 / 2 {\displaystyle \alpha =1/2} , it corresponds to the diameter and the s i s c {\displaystyle sisc} is π 2 / 4 {\displaystyle \pi ^{2}/4} as stated above.</li> <li>Now, assume that you have 2 n {\displaystyle 2^{n}} points on a circle with radius 2 n / π {\displaystyle 2^{n}/\pi } and center O {\displaystyle O} , and 2 n + 1 {\displaystyle 2^{n+1}} points on a circle with radius 2 n + 1 / π {\displaystyle 2^{n+1}/\pi } and center R {\displaystyle R} . The induction step consists in showing that these 2 circles have the same s i s c {\displaystyle sisc} for a given α {\displaystyle \alpha } .</li></ul> <ul><li>Start by drawing the circles so that they share point Q {\displaystyle Q} . Note that R {\displaystyle R} lies on the smaller circle. Then, note that 2 n + 1 {\displaystyle 2^{n+1}} is always even, and a simple geometric argument shows that you can pick <i>pairs</i> of opposite points P 1 {\displaystyle P_{1}} and P 2 {\displaystyle P_{2}} on the larger circle by joining each pair with a diameter. Furthermore, for each pair, one of the points will be in the "lower" half of the circle (closer to Q {\displaystyle Q} ) and the other in the "upper" half.</li></ul> <ul><li>The diameter of the bigger circle P 1 P 2 {\displaystyle P_{1}P_{2}} cuts the smaller circle at R {\displaystyle R} and at another point P {\displaystyle P} . You can then make the following considerations: <ul><li> P 1 Q ^ P 2 {\displaystyle P_{1}{\widehat {Q}}P_{2}} is a right angle, since P 1 P 2 {\displaystyle P_{1}P_{2}} is a diameter.</li> <li> Q P ^ R {\displaystyle Q{\widehat {P}}R} is a right angle, since Q R {\displaystyle QR} is a diameter.</li> <li> Q R ^ P 2 = Q R ^ P {\displaystyle Q{\widehat {R}}P_{2}=Q{\widehat {R}}P} is half of Q O ^ P {\displaystyle Q{\widehat {O}}P} for the <a href="/facts/Inscribed_angle/fTYeVKxt">Inscribed Angle Theorem</a>.</li> <li>Hence, the arc Q P {\displaystyle QP} is equal to the arc Q P 2 {\displaystyle QP_{2}} , again because the radius is half.</li> <li>The chord Q P {\displaystyle QP} is the height of the right triangle Q P 1 P 2 {\displaystyle QP_{1}P_{2}} , hence for the Inverse Pythagorean Theorem:</li></ul></li></ul> <p> 1 Q P ¯ 2 = 1 Q P 1 ¯ 2 + 1 Q P 2 ¯ 2 {\displaystyle {\frac {1}{{\overline {QP}}^{2}}}={\frac {1}{{\overline {QP_{1}}}^{2}}}+{\frac {1}{{\overline {QP_{2}}}^{2}}}} </p> <ul><li>Hence for half of the points on the bigger circle (the ones in the lower half) there is a corresponding point on the smaller circle with the same arc distance from Q {\displaystyle Q} (since the circumference of the smaller circle is half the one of the bigger circle, the last two points closer to R {\displaystyle R} must have arc distance 2 as well). Vice versa, for each of the 2 n {\displaystyle 2^{n}} points on the smaller circle, we can build a pair of points on the bigger circle, and all of these points are equidistant and have the same arc distance from Q {\displaystyle Q} .</li> <li>Furthermore, the total s i s c {\displaystyle sisc} for the bigger circle is the same as the s i s c {\displaystyle sisc} for the smaller circle, since each pair of points on the bigger circle has the same inverse square sum as the corresponding point on the smaller circle.<a class="footnote-ref" id="fnref:14" href="#fn:14"><sup>14</sup></a></li></ul> <h2 id="other-identities">Other identities</h2> <p>See the special cases of the identities for the <a href="/facts/Riemann_zeta_function/MnfRu7lY">Riemann zeta function</a> when s = 2. {\displaystyle s=2.} Other notably special identities and representations of this constant appear in the sections below. </p> <h3>Series representations</h3> <p>The following are series representations of the constant:<a class="footnote-ref" id="fnref:15" href="#fn:15"><sup>15</sup></a> ζ ( 2 ) = 3 ∑ k = 1 ∞ 1 k 2 ( 2 k k ) = ∑ i = 1 ∞ ∑ j = 1 ∞ ( i − 1 ) ! ( j − 1 ) ! ( i + j ) ! . {\displaystyle {\begin{aligned}\zeta (2)&=3\sum _{k=1}^{\infty }{\frac {1}{k^{2}{\binom {2k}{k}}}}\\[6pt]&=\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }{\frac {(i-1)!(j-1)!}{(i+j)!}}.\end{aligned}}} </p><p>There are also <a href="/facts/Bailey%25E2%2580%2593Borwein%25E2%2580%2593Plouffe_formula/m4v34n9d">BBP-type</a> series expansions for <i>ζ</i>(2).<a class="footnote-ref" id="fnref:16" href="#fn:16"><sup>16</sup></a> </p> <h3>Integral representations</h3> <p>The following are integral representations of ζ ( 2 ) : {\displaystyle \zeta (2){\text{:}}} <a class="footnote-ref" id="fnref:17" href="#fn:17"><sup>17</sup></a><a class="footnote-ref" id="fnref:18" href="#fn:18"><sup>18</sup></a><a class="footnote-ref" id="fnref:19" href="#fn:19"><sup>19</sup></a> ζ ( 2 ) = − ∫ 0 1 log x 1 − x d x = ∫ 0 ∞ x e x − 1 d x = ∫ 0 1 ( log x ) 2 ( 1 + x ) 2 d x = 2 + 2 ∫ 1 ∞ ⌊ x ⌋ − x x 3 d x = exp ( 2 ∫ 2 ∞ π ( x ) x ( x 2 − 1 ) d x ) = ∫ 0 1 ∫ 0 1 d x d y 1 − x y = 4 3 ∫ 0 1 ∫ 0 1 d x d y 1 − ( x y ) 2 = ∫ 0 1 ∫ 0 1 1 − x 1 − x y d x d y + 2 3 . {\displaystyle {\begin{aligned}\zeta (2)&=-\int _{0}^{1}{\frac {\log x}{1-x}}\,dx\\[6pt]&=\int _{0}^{\infty }{\frac {x}{e^{x}-1}}\,dx\\[6pt]&=\int _{0}^{1}{\frac {(\log x)^{2}}{(1+x)^{2}}}\,dx\\[6pt]&=2+2\int _{1}^{\infty }{\frac {\lfloor x\rfloor -x}{x^{3}}}\,dx\\[6pt]&=\exp \left(2\int _{2}^{\infty }{\frac {\pi (x)}{x(x^{2}-1)}}\,dx\right)\\[6pt]&=\int _{0}^{1}\int _{0}^{1}{\frac {dx\,dy}{1-xy}}\\[6pt]&={\frac {4}{3}}\int _{0}^{1}\int _{0}^{1}{\frac {dx\,dy}{1-(xy)^{2}}}\\[6pt]&=\int _{0}^{1}\int _{0}^{1}{\frac {1-x}{1-xy}}\,dx\,dy+{\frac {2}{3}}.\end{aligned}}} </p> <h3>Continued fractions</h3> <p>In van der Poorten's classic article chronicling <a href="/facts/Ap%25C3%25A9ry%2527s_constant/59do2TbQ">Apéry's proof of the irrationality of ζ ( 3 ) {\displaystyle \zeta (3)} </a>,<a class="footnote-ref" id="fnref:20" href="#fn:20"><sup>20</sup></a> the author notes as "a red herring" the similarity of a <a href="/facts/Simple_continued_fraction/Wg1cMPxN">simple continued fraction</a> for Apery's constant, and the following one for the Basel constant: ζ ( 2 ) 5 = 1 v ~ 1 − 1 4 v ~ 2 − 2 4 v ~ 3 − 3 4 v ~ 4 − ⋱ , {\displaystyle {\frac {\zeta (2)}{5}}={\cfrac {1}{{\widetilde {v}}_{1}-{\cfrac {1^{4}}{{\widetilde {v}}_{2}-{\cfrac {2^{4}}{{\widetilde {v}}_{3}-{\cfrac {3^{4}}{{\widetilde {v}}_{4}-\ddots }}}}}}}},} where v ~ n = 11 n 2 − 11 n + 3 ↦ { 3 , 25 , 69 , 135 , … } {\displaystyle {\widetilde {v}}_{n}=11n^{2}-11n+3\mapsto \{3,25,69,135,\ldots \}} . Another continued fraction of a similar form is:<a class="footnote-ref" id="fnref:21" href="#fn:21"><sup>21</sup></a> ζ ( 2 ) 2 = 1 v 1 − 1 4 v 2 − 2 4 v 3 − 3 4 v 4 − ⋱ , {\displaystyle {\frac {\zeta (2)}{2}}={\cfrac {1}{v_{1}-{\cfrac {1^{4}}{v_{2}-{\cfrac {2^{4}}{v_{3}-{\cfrac {3^{4}}{v_{4}-\ddots }}}}}}}},} where v n = 2 n − 1 ↦ { 1 , 3 , 5 , 7 , 9 , … } {\displaystyle v_{n}=2n-1\mapsto \{1,3,5,7,9,\ldots \}} . </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/List_of_sums_of_reciprocals/B2eAx0vL">List of sums of reciprocals</a></li></ul> <ul><li><a href="/facts/Andr%25C3%25A9_Weil/Q4Dd0Gr4">Weil, André</a> (1983), <i>Number Theory: An Approach Through History</i>, Springer-Verlag, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-8176-3141-0.</li> <li><a href="/facts/William_Dunham_(mathematician)/CH7qquB0">Dunham, William</a> (1999), <a href="https://archive.org/details/eulermasterofusa0000dunh"><i>Euler: The Master of Us All</i></a>, <a href="/facts/Mathematical_Association_of_America/G3BkG5YX">Mathematical Association of America</a>, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-88385-328-0.</li> <li><a href="/facts/John_Derbyshire/efG5wLFS">Derbyshire, John</a> (2003), <a href="https://archive.org/details/primeobsessionbe00derb_0"><i>Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics</i></a>, Joseph Henry Press, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-309-08549-7.</li> <li><a href="/facts/Harold_Edwards_(mathematician)/JKfNgfJB">Edwards, Harold M.</a> (2001), <i>Riemann's Zeta Function</i>, Dover, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-486-41740-9.</li></ul> <h2 id="notes">Notes</h2> <h2 id="external-links">External links</h2> <ul><li><a href="https://plus.maths.org/content/infinite-series-surprises">An infinite series of surprises</a> by C. J. Sangwin</li> <li><a href="https://fylux.github.io/2017/03/30/Pi/">From <i>ζ</i>(2) to Π. The Proof.</a> step-by-step proof</li> <li><a href="http://eulerarchive.maa.org//docs/translations/E352.pdf"><i>Remarques sur un beau rapport entre les series des puissances tant directes que reciproques</i></a> (PDF), English translation with notes of Euler's paper by Lucas Willis and Thomas J. Osler</li> <li>Ed Sandifer, <a href="http://eulerarchive.maa.org/hedi/HEDI-2003-12.pdf"><i>How Euler did it</i></a> (PDF)</li> <li>James A. Sellers (February 5, 2002), <a href="http://www.personal.psu.edu/jxs23/p25.pdf"><i>Beyond Mere Convergence</i></a> (PDF), retrieved 2004-02-27</li> <li>Robin Chapman, <a href="http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf"><i>Evaluating</i> <i>ζ</i>(2)</a> (fourteen proofs)</li> <li><a href="https://web.archive.org/web/20181130113058/https://giphy.com/gifs/math-visualization-algorithm-xThuW9Pyh8jXvfbrUc">Visualization of Euler's factorization of the sine function</a></li> <li>Johan W Ästlund (December 8, 2010), <a href="http://www.math.chalmers.se/~wastlund/Cosmic.pdf"><i>Summing inverse squares by Euclidean geometry</i></a> (PDF) <ul><li><a href="https://www.youtube.com/watch?v=d-o3eB9sfls">Why is pi here? And why is it squared? A geometric answer to the Basel problem</a> on <a href="/facts/YouTube_video_(identifier)/db276jst">YouTube</a> (animated proof based on the above)</li></ul></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Ayoub, Raymond (1974), "Euler and the zeta function", Amer. Math. Monthly, 81 (10): 1067–86, doi:10.2307/2319041, JSTOR 2319041, archived from the original on 2019-08-14, retrieved 2021-01-25 <a href="https://web.archive.org/web/20190814233022/https://www.maa.org/programs/maa-awards/writing-awards/euler-and-the-zeta-function" target="_blank">https://web.archive.org/web/20190814233022/https://www.maa.org/programs/maa-awards/writing-awards/euler-and-the-zeta-function</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>E41 – De summis serierum reciprocarum <a href="https://scholarlycommons.pacific.edu/euler-works/41/" target="_blank">https://scholarlycommons.pacific.edu/euler-works/41/</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>Sloane, N. J. A. (ed.), "Sequence A013661", The On-Line Encyclopedia of Integer Sequences, OEIS Foundation <a href="/wiki/Neil_Sloane" target="_blank">/wiki/Neil_Sloane</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>Vandervelde, Sam (2009), "Chapter 9: Sneaky segments", Circle in a Box, MSRI Mathematical Circles Library, Mathematical Sciences Research Institute and American Mathematical Society, pp. 101–106 <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>A priori, since the left-hand-side is a polynomial (of infinite degree) we can write it as a product of its roots as sin ( x ) = x ( x 2 − π 2 ) ( x 2 − 4 π 2 ) ( x 2 − 9 π 2 ) ⋯ = A x ( 1 − x 2 π 2 ) ( 1 − x 2 4 π 2 ) ( 1 − x 2 9 π 2 ) ⋯ . {\displaystyle {\begin{aligned}\sin(x)&=x(x^{2}-\pi ^{2})(x^{2}-4\pi ^{2})(x^{2}-9\pi ^{2})\cdots \\&=Ax\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots .\end{aligned}}} Then since we know from elementary calculus that lim x → 0 sin ( x ) x = 1 {\displaystyle \lim _{x\rightarrow 0}{\frac {\sin(x)}{x}}=1} , we conclude that the leading constant must satisfy A = 1 {\displaystyle A=1} . <a href="/wiki/Polynomial" target="_blank">/wiki/Polynomial</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> <li id="fn:6"><p>In particular, letting H n ( 2 ) := ∑ k = 1 n k − 2 {\displaystyle H_{n}^{(2)}:=\sum _{k=1}^{n}k^{-2}} denote a generalized second-order harmonic number, we can easily prove by induction that [ x 2 ] ∏ k = 1 n ( 1 − x 2 π 2 ) = − H n ( 2 ) π 2 → − ζ ( 2 ) π 2 {\displaystyle [x^{2}]\prod _{k=1}^{n}\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)=-{\frac {H_{n}^{(2)}}{\pi ^{2}}}\rightarrow -{\frac {\zeta (2)}{\pi ^{2}}}} as n → ∞ {\displaystyle n\rightarrow \infty } . <a href="/wiki/Generalized_harmonic_number" target="_blank">/wiki/Generalized_harmonic_number</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li> <li id="fn:7"><p>Havil, J. (2003), Gamma: Exploring Euler's Constant, Princeton, New Jersey: Princeton University Press, pp. 37–42 (Chapter 4), ISBN 0-691-09983-9 <a href="0-691-09983-9" target="_blank">0-691-09983-9</a> <a href="#fnref:7" class="footnote-back-ref">↩</a></p></li> <li id="fn:8"><p>Cf., the formulae for generalized Stirling numbers proved in: Schmidt, M. D. (2018), "Combinatorial Identities for Generalized Stirling Numbers Expanding f-Factorial Functions and the f-Harmonic Numbers", J. Integer Seq., 21 (Article 18.2.7) <a href="https://cs.uwaterloo.ca/journals/JIS/VOL21/Schmidt/schmidt18.html" target="_blank">https://cs.uwaterloo.ca/journals/JIS/VOL21/Schmidt/schmidt18.html</a> <a href="#fnref:8" class="footnote-back-ref">↩</a></p></li> <li id="fn:9"><p>Arakawa, Tsuneo; Ibukiyama, Tomoyoshi; Kaneko, Masanobu (2014), Bernoulli Numbers and Zeta Functions, Springer, p. 61, ISBN 978-4-431-54919-2 <a href="978-4-431-54919-2" target="_blank">978-4-431-54919-2</a> <a href="#fnref:9" class="footnote-back-ref">↩</a></p></li> <li id="fn:10"><p>Freitas, F. L. (2023), "Solution of the Basel problem using the Feynman integral trick", arXiv:2312.04608 [math.CA] <a href="/wiki/ArXiv_(identifier)" target="_blank">/wiki/ArXiv_(identifier)</a> <a href="#fnref:10" class="footnote-back-ref">↩</a></p></li> <li id="fn:11"><p>Ransford, T J (Summer 1982), "An Elementary Proof of ∑ 1 ∞ 1 n 2 = π 2 6 {\displaystyle \sum _{1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}} " (PDF), Eureka, 42 (1): 3–4, archived from the original (PDF) on June 10, 2020 <a href="https://web.archive.org/web/20200610100509/https://www.archim.org.uk/eureka/archive/Eureka-42.pdf" target="_blank">https://web.archive.org/web/20200610100509/https://www.archim.org.uk/eureka/archive/Eureka-42.pdf</a> <a href="#fnref:11" class="footnote-back-ref">↩</a></p></li> <li id="fn:12"><p>Aigner, Martin; Ziegler, Günter M. (2001), Proofs from THE BOOK (2nd ed.), Springer, p. 32, ISBN 9783662043158; this anecdote is missing from later editions of this book, which replace it with earlier history of the same proof. <a href="9783662043158" target="_blank">9783662043158</a> <a href="#fnref:12" class="footnote-back-ref">↩</a></p></li> <li id="fn:13"><p>Vladimir Platonov; Andrei Rapinchuk (1994), Algebraic groups and number theory, translated by Rachel Rowen, Academic Press| <a href="/wiki/Vladimir_Platonov" target="_blank">/wiki/Vladimir_Platonov</a> <a href="#fnref:13" class="footnote-back-ref">↩</a></p></li> <li id="fn:14"><p>Johan Wästlund (December 8, 2010). "Summing Inverse Squares by Euclidean Geometry" (PDF). Chalmers University of Technology. Department of Mathematics, Chalmers University. Retrieved 2024-10-11. <a href="https://www.math.chalmers.se/~wastlund/Cosmic.pdf" target="_blank">https://www.math.chalmers.se/~wastlund/Cosmic.pdf</a> <a href="#fnref:14" class="footnote-back-ref">↩</a></p></li> <li id="fn:15"><p>Weisstein, Eric W., "Riemann Zeta Function \zeta(2)", MathWorld <a href="/wiki/Eric_W._Weisstein" target="_blank">/wiki/Eric_W._Weisstein</a> <a href="#fnref:15" class="footnote-back-ref">↩</a></p></li> <li id="fn:16"><p>Weisstein, Eric W., "Riemann Zeta Function \zeta(2)", MathWorld <a href="/wiki/Eric_W._Weisstein" target="_blank">/wiki/Eric_W._Weisstein</a> <a href="#fnref:16" class="footnote-back-ref">↩</a></p></li> <li id="fn:17"><p>Connon, D. F. (2007), "Some series and integrals involving the Riemann zeta function, binomial coefficients and the harmonic numbers (Volume I)", arXiv:0710.4022 [math.HO] <a href="/wiki/ArXiv_(identifier)" target="_blank">/wiki/ArXiv_(identifier)</a> <a href="#fnref:17" class="footnote-back-ref">↩</a></p></li> <li id="fn:18"><p>Weisstein, Eric W., "Double Integral", MathWorld <a href="/wiki/Eric_W._Weisstein" target="_blank">/wiki/Eric_W._Weisstein</a> <a href="#fnref:18" class="footnote-back-ref">↩</a></p></li> <li id="fn:19"><p>Weisstein, Eric W., "Hadjicostas's Formula", MathWorld <a href="/wiki/Eric_W._Weisstein" target="_blank">/wiki/Eric_W._Weisstein</a> <a href="#fnref:19" class="footnote-back-ref">↩</a></p></li> <li id="fn:20"><p>van der Poorten, Alfred (1979), "A proof that Euler missed ... Apéry's proof of the irrationality of ζ(3)" (PDF), The Mathematical Intelligencer, 1 (4): 195–203, doi:10.1007/BF03028234, S2CID 121589323, archived from the original (PDF) on 2011-07-06 <a href="/wiki/Alfred_van_der_Poorten" target="_blank">/wiki/Alfred_van_der_Poorten</a> <a href="#fnref:20" class="footnote-back-ref">↩</a></p></li> <li id="fn:21"><p>Berndt, Bruce C. (1989), Ramanujan's Notebooks: Part II, Springer-Verlag, p. 150, ISBN 978-0-387-96794-3 <a href="978-0-387-96794-3" target="_blank">978-0-387-96794-3</a> <a href="#fnref:21" class="footnote-back-ref">↩</a></p></li> </ol>