# Limit comparison test

In mathematics, 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

Suppose that we have two series ${\displaystyle \Sigma _a_}$ and ${\displaystyle \Sigma _b_}$ with ${\displaystyle a_\geq 0,b_>0}$ for all ${\displaystyle n}$.

Then if ${\displaystyle \lim _{\frac }}}=c}$ with ${\displaystyle 0, then either both series converge or both series diverge.[1]

## Proof

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

${\displaystyle -\varepsilon <{\frac }}}-c<\varepsilon }$
${\displaystyle c-\varepsilon <{\frac }}}
${\displaystyle (c-\varepsilon )b_

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

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

That is, both series converge or both series diverge.

## Example

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

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

## One-sided version

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

## Example

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

## Converse of the one-sided comparison test

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

## Example

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