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 ${\mathrm{\Sigma }}_n{a}_n$ and ${\mathrm{\Sigma }}_n{b}_n$ with $a_n\ge 0,b_n>0$ for all $n$. Then if $\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=c$ with $0<c<\mathrm{\infty }$, then either both series converge or both series diverge.

Proof

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

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

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

$(c-\epsilon )b_n<a_n<(c+\epsilon )b_n$

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

Similarly $a_n<(c+\epsilon )b_n$, so if $\sum _n{a}_n$ diverges, again by the direct comparison test, so does $\sum _n{b}_n$.

That is, both series converge or both series diverge.

Example

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

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

One-sided version

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

Example

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

Converse of the one-sided comparison test

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

Example

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

See also

• Convergence tests
• Direct comparison test

Further reading

• Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011, Template:Isbn, 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

• Pauls Online Notes on Comparison Test