Limit comparison test

Proof

Because ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=c}$ we know that for every ${\displaystyle \epsilon >0}$ there is a positive integer ${\displaystyle n_0}$ such that for all ${\displaystyle n\ge n_0}$ we have that ${\displaystyle \left|\frac{a_n}{b_n}-c\right|<\epsilon }$, or equivalently

${\displaystyle -\epsilon <\frac{a_n}{b_n}-c<\epsilon }$

${\displaystyle c-\epsilon <\frac{a_n}{b_n}<c+\epsilon }$

${\displaystyle (c-\epsilon )b_n<a_n<(c+\epsilon )b_n}$

As ${\displaystyle c>0}$ we can choose ${\displaystyle \epsilon }$ to be sufficiently small such that ${\displaystyle c-\epsilon }$ is positive. So ${\displaystyle b_n<\frac{1}{c-\epsilon }{a}_n}$ and by the direct comparison test, if ${\displaystyle \sum _n{a}_n}$ converges then so does ${\displaystyle \sum _n{b}_n}$.

Similarly ${\displaystyle a_n<(c+\epsilon )b_n}$, 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

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

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

One-sided version

One can state a one-sided comparison test by using limit superior. Let ${\displaystyle a_n,b_n\ge 0}$ for all ${\displaystyle n}$. Then if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}\frac{a_n}{b_n}=c}$ with ${\displaystyle 0\le c<\mathrm{\infty }}$ and ${\displaystyle {\mathrm{\Sigma }}_n{b}_n}$ converges, necessarily ${\displaystyle {\mathrm{\Sigma }}_n{a}_n}$ converges.

Example

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 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{lim}(1-(-1{)}^n)}$ does not exist, so we cannot apply the standard comparison test. However, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{lim sup}(1-(-1{)}^n)=2\in [0,\mathrm{\infty })}$ and since ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}}$ converges, the one-sided comparison test implies that ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1-(-1{)}^n}{n^2}}$ converges.

Converse of the one-sided comparison test

Let ${\displaystyle a_n,b_n\ge 0}$ for all ${\displaystyle n}$. If ${\displaystyle {\mathrm{\Sigma }}_n{a}_n}$ diverges and ${\displaystyle {\mathrm{\Sigma }}_n{b}_n}$ converges, then necessarily ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}\frac{a_n}{b_n}=\mathrm{\infty }}$, that is, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim inf}\frac{b_n}{a_n}=0}$. The essential content here is that in some sense the numbers ${\displaystyle a_n}$ are larger than the numbers ${\displaystyle b_n}$.

Example

Let ${\displaystyle f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n}$ 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 ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}n|a_n{|}^2}$. Moreover, ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}1/n}$ diverges. Therefore, by the converse of the comparison test, we have ${\displaystyle \underset{n\to \mathrm{\infty }}{lim inf}\frac{n|a_n{|}^2}{1/n}=\underset{n\to \mathrm{\infty }}{lim inf}(n|a_n|{)}^2=0}$, that is, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim inf}n|a_n|=0}$.