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Limit comparison test

Method of testing for the convergence of an infinite series

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Inmathematics, the limit comparison test (LCT) (in contrast with the related direct comparison test) is a method of testing for the convergence of an infinite series.

Statement[edit]

Suppose that we have two series ${\displaystyle \Sigma _{n}a_{n}}$ and ${\displaystyle \Sigma _{n}b_{n}}$ with ${\displaystyle a_{n}\geq 0,b_{n}>0}$ for all ${\displaystyle n}$. Then if ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ with ${\displaystyle 0<c<\infty }$, then either both series converge or both series diverge.^[1]

Proof[edit]

Because ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ we know that for every ${\displaystyle \varepsilon >0}$ there is a positive integer ${\displaystyle n_{0}}$ such that for all ${\displaystyle n\geq n_{0}}$ we have that ${\displaystyle \left|{\frac {a_{n}}{b_{n}}}-c\right|<\varepsilon }$, or equivalently

${\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon }$

${\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}<c+\varepsilon }$

${\displaystyle (c-\varepsilon )b_{n}<a_{n}<(c+\varepsilon )b_{n}}$

As${\displaystyle c>0}$ we can choose ${\displaystyle \varepsilon }$ to be sufficiently small such that ${\displaystyle c-\varepsilon }$ is positive. So ${\displaystyle b_{n}<{\frac {1}{c-\varepsilon }}a_{n}}$ and by the direct comparison test, if ${\displaystyle \sum _{n}a_{n}}$ converges then so does ${\displaystyle \sum _{n}b_{n}}$.

Similarly ${\displaystyle a_{n}<(c+\varepsilon )b_{n}}$, so if ${\displaystyle \sum _{n}a_{n}}$ diverges, again by the direct comparison test, so does ${\displaystyle \sum _{n}b_{n}}$.

That is, both series converge or both series diverge.

Example[edit]

We want to determine if the series ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}+2n}}}$ converges. For this we compare it with the convergent series ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}$

As${\displaystyle \lim _{n\to \infty }{\frac {1}{n^{2}+2n}}{\frac {n^{2}}{1}}=1>0}$ we have that the original series also converges.

One-sided version[edit]

One can state a one-sided comparison test by using limit superior. Let ${\displaystyle a_{n},b_{n}\geq 0}$ for all ${\displaystyle n}$. Then if ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ with ${\displaystyle 0\leq c<\infty }$ and ${\displaystyle \Sigma _{n}b_{n}}$ converges, necessarily ${\displaystyle \Sigma _{n}a_{n}}$ converges.

Example[edit]

Let ${\displaystyle a_{n}={\frac {1-(-1)^{n}}{n^{2}}}}$ and ${\displaystyle b_{n}={\frac {1}{n^{2}}}}$ for all natural numbers ${\displaystyle n}$. Now ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\to \infty }(1-(-1)^{n})}$ does not exist, so we cannot apply the standard comparison test. However, ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\limsup _{n\to \infty }(1-(-1)^{n})=2\in [0,\infty )}$ and since ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$ converges, the one-sided comparison test implies that ${\displaystyle \sum _{n=1}^{\infty }{\frac {1-(-1)^{n}}{n^{2}}}}$ converges.

Converse of the one-sided comparison test[edit]

Let ${\displaystyle a_{n},b_{n}\geq 0}$ for all ${\displaystyle n}$. If ${\displaystyle \Sigma _{n}a_{n}}$ diverges and ${\displaystyle \Sigma _{n}b_{n}}$ converges, then necessarily ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\infty }$, that is, ${\displaystyle \liminf _{n\to \infty }{\frac {b_{n}}{a_{n}}}=0}$. The essential content here is that in some sense the numbers ${\displaystyle a_{n}}$ are larger than the numbers ${\displaystyle b_{n}}$.

Example[edit]

Let ${\displaystyle f($z)=\sum _{n=0}^{\infty }a_{n}z^{n}} be analytic in the unit disc ${\displaystyle D=\{z\in \mathbb {C} :|z|<1\}}$ and have image of finite area. By Parseval's formula the area of the image of ${\displaystyle f}$ is proportional to ${\displaystyle \sum _{n=1}^{\infty }n|a_{n}|^{2}}$. Moreover, ${\displaystyle \sum _{n=1}^{\infty }1/n}$ diverges. Therefore, by the converse of the comparison test, we have ${\displaystyle \liminf _{n\to \infty }{\frac {n|a_{n}|^{2}}{1/n}}=\liminf _{n\to \infty }(n|a_{n}|)^{2}=0}$, that is, ${\displaystyle \liminf _{n\to \infty }n|a_{n}|=0}$.

See also[edit]

• Convergence tests
• Direct comparison test

References[edit]

Further reading[edit]

• Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011, ISBN 9780817682897, pp. 50
• Michele Longo and Vincenzo Valori: The Comparison Test: Not Just for Nonnegative Series. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210 (JSTOR)
• J. Marshall Ash: The Limit Comparison Test Needs Positivity. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375 (JSTOR)

External links[edit]

• Pauls Online Notes on Comparison Test