# 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 $\Sigma _a_$ and $\Sigma _b_$ with $a_\geq 0,b_>0$ for all $n$ .

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

## Proof

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

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

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

That is, both series converge or both series diverge.

## Example

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

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

## One-sided version

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

## Example

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

## Converse of the one-sided comparison test

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

## Example

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