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 $$ a_n\geq 0, b_n > 0 $$ 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 every $$ \varepsilon > 0 $$ 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 $$ \sum_n a_n $$ 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 $$ \sum_{n=1}^{\infty} \frac{1}{n^2 + 2n} $$ converges. For this we compare it 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 $$ f(z)=\sum_{n=0}^{\infty}a_nz^n $$ 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 proportional to $$ \sum_{n=1}^{\infty} n|a_n|^2$$. 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, $$ \liminf_{n\to\infty} n|a_n| = 0 $$.