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Limit comparison test
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<h2 id="statement">Statement</h2> <p>Suppose that we have two series Σ n a n {\displaystyle \Sigma _{n}a_{n}} and Σ n b n {\displaystyle \Sigma _{n}b_{n}} with a n ≥ 0 , b n > 0 {\displaystyle a_{n}\geq 0,b_{n}>0} for all n {\displaystyle n} . Then if lim n → ∞ a n b n = c {\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c} with 0 < c < ∞ {\displaystyle 0<c<\infty } , then either both series converge or both series diverge.<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> </p> <h2 id="proof">Proof</h2> <p>Because lim n → ∞ a n b n = c {\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c} we know that for every ε > 0 {\displaystyle \varepsilon >0} there is a positive integer n 0 {\displaystyle n_{0}} such that for all n ≥ n 0 {\displaystyle n\geq n_{0}} we have that | a n b n − c | < ε {\displaystyle \left|{\frac {a_{n}}{b_{n}}}-c\right|<\varepsilon } , or equivalently </p> − ε < a n b n − c < ε {\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon } c − ε < a n b n < c + ε {\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}<c+\varepsilon } ( c − ε ) b n < a n < ( c + ε ) b n {\displaystyle (c-\varepsilon )b_{n}<a_{n}<(c+\varepsilon )b_{n}} <p>As c > 0 {\displaystyle c>0} we can choose ε {\displaystyle \varepsilon } to be sufficiently small such that c − ε {\displaystyle c-\varepsilon } is positive. So b n < 1 c − ε a n {\displaystyle b_{n}<{\frac {1}{c-\varepsilon }}a_{n}} and by the <a href="/facts/Direct_comparison_test/AeHYrl7o">direct comparison test</a>, if ∑ n a n {\displaystyle \sum _{n}a_{n}} converges then so does ∑ n b n {\displaystyle \sum _{n}b_{n}} . </p><p>Similarly a n < ( c + ε ) b n {\displaystyle a_{n}<(c+\varepsilon )b_{n}} , so if ∑ n b n {\displaystyle \sum _{n}b_{n}} diverges, again by the direct comparison test, so does ∑ n a n {\displaystyle \sum _{n}a_{n}} . </p><p>That is, both series converge or both series diverge. </p> <h2 id="example">Example</h2> <p>We want to determine if the series ∑ n = 1 ∞ 1 n 2 + 2 n {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}+2n}}} converges. For this we compare it with the convergent series ∑ n = 1 ∞ 1 n 2 = π 2 6 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}} </p><p>As lim n → ∞ 1 n 2 + 2 n n 2 1 = 1 > 0 {\displaystyle \lim _{n\to \infty }{\frac {1}{n^{2}+2n}}{\frac {n^{2}}{1}}=1>0} we have that the original series also converges. </p> <h2 id="one-sided-version">One-sided version</h2> <p>One can state a one-sided comparison test by using <a href="/facts/Limit_superior/QeAhlygs">limit superior</a>. Let a n , b n ≥ 0 {\displaystyle a_{n},b_{n}\geq 0} for all n {\displaystyle n} . Then if lim sup n → ∞ a n b n = c {\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c} with 0 ≤ c < ∞ {\displaystyle 0\leq c<\infty } and Σ n b n {\displaystyle \Sigma _{n}b_{n}} converges, necessarily Σ n a n {\displaystyle \Sigma _{n}a_{n}} converges. </p> <h2 id="example">Example</h2> <p>Let a n = 1 − ( − 1 ) n n 2 {\displaystyle a_{n}={\frac {1-(-1)^{n}}{n^{2}}}} and b n = 1 n 2 {\displaystyle b_{n}={\frac {1}{n^{2}}}} for all natural numbers n {\displaystyle n} . Now lim n → ∞ a n b n = lim n → ∞ ( 1 − ( − 1 ) n ) {\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\to \infty }(1-(-1)^{n})} does not exist, so we cannot apply the standard comparison test. However, lim sup n → ∞ a n b n = lim sup n → ∞ ( 1 − ( − 1 ) n ) = 2 ∈ [ 0 , ∞ ) {\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\limsup _{n\to \infty }(1-(-1)^{n})=2\in [0,\infty )} and since ∑ n = 1 ∞ 1 n 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}} converges, the one-sided comparison test implies that ∑ n = 1 ∞ 1 − ( − 1 ) n n 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1-(-1)^{n}}{n^{2}}}} converges. </p> <h2 id="converse-of-the-one-sided-comparison-test">Converse of the one-sided comparison test</h2> <p>Let a n , b n ≥ 0 {\displaystyle a_{n},b_{n}\geq 0} for all n {\displaystyle n} . If Σ n a n {\displaystyle \Sigma _{n}a_{n}} diverges and Σ n b n {\displaystyle \Sigma _{n}b_{n}} converges, then necessarily lim sup n → ∞ a n b n = ∞ {\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\infty } , that is, lim inf n → ∞ b n a n = 0 {\displaystyle \liminf _{n\to \infty }{\frac {b_{n}}{a_{n}}}=0} . The essential content here is that in some sense the numbers a n {\displaystyle a_{n}} are larger than the numbers b n {\displaystyle b_{n}} . </p> <h2 id="example">Example</h2> <p>Let f ( z ) = ∑ n = 0 ∞ a n z n {\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}} be analytic in the unit disc D = { z ∈ C : | z | < 1 } {\displaystyle D=\{z\in \mathbb {C} :|z|<1\}} and have image of finite area. By <a href="/facts/Parseval%2527s_formula/EukqvgUp">Parseval's formula</a> the area of the image of f {\displaystyle f} is proportional to ∑ n = 1 ∞ n | a n | 2 {\displaystyle \sum _{n=1}^{\infty }n|a_{n}|^{2}} . Moreover, ∑ n = 1 ∞ 1 / n {\displaystyle \sum _{n=1}^{\infty }1/n} diverges. Therefore, by the converse of the comparison test, we have lim inf n → ∞ n | a n | 2 1 / n = lim inf n → ∞ ( n | a n | ) 2 = 0 {\displaystyle \liminf _{n\to \infty }{\frac {n|a_{n}|^{2}}{1/n}}=\liminf _{n\to \infty }(n|a_{n}|)^{2}=0} , that is, lim inf n → ∞ n | a n | = 0 {\displaystyle \liminf _{n\to \infty }n|a_{n}|=0} . </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Convergence_tests/kSwGCpTe">Convergence tests</a></li> <li><a href="/facts/Direct_comparison_test/AeHYrl7o">Direct comparison test</a></li></ul> <h2 id="further-reading">Further reading</h2> <ul><li>Rinaldo B. Schinazi: <i>From Calculus to Analysis</i>. Springer, 2011, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 9780817682897, pp. <a href="https://books.google.com/books?id=VybcUbhGvjsC&pg=PA50">50</a></li> <li>Michele Longo and Vincenzo Valori: <i>The Comparison Test: Not Just for Nonnegative Series</i>. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210 (<a href="https://www.jstor.org/stable/27642937">JSTOR</a>)</li> <li>J. Marshall Ash: <i>The Limit Comparison Test Needs Positivity</i>. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375 (<a href="https://www.jstor.org/stable/pdf/10.4169/math.mag.85.5.374.pdf">JSTOR</a>)</li></ul> <h2 id="external-links">External links</h2> <ul><li><a href="http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx">Pauls Online Notes on Comparison Test </a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Swokowski, Earl (1983), Calculus with analytic geometry (Alternate ed.), Prindle, Weber & Schmidt, p. 516, ISBN 0-87150-341-7 <a href="0-87150-341-7" target="_blank">0-87150-341-7</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>