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 $\Sigma_n a_n$ and $\Sigma_n b_n$ with $a_n, b_n \geq 0$ for all $n$.

Then if $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ with $0 < c < \infty$ then either both series converge or both series diverge.

Proof

Because $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ we know that for all $\varepsilon >0$ there is a positive integer $n_{0}$ such that for all $n \geq n_0$ we have that $\left| \frac{a_n}{b_n} - c \right| < \varepsilon$, or equivalently

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

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

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

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

Similarly $a_n < (c + \varepsilon)b_n$, so if $\sum_n b_n$ converges, again by the direct comparison test, so does $\sum_n a_n$.

That is both series converge or both series diverge.

Example

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

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

One-sided version

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

Example

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

Converse of the one-sided comparison test

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

Example

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

See also

• Convergence tests
• Direct comparison test

External links

• Pauls Online Notes on Comparison Test