Proof of Limit Comparison Test

Statement of Test

Let $(a_n)_{n=1}^{\infty},(b_n)_{n=1}^{\infty}$ be two strictly positive sequences of real numbers such that $\frac{a_n}{b_n} \to L$ as $n \to \infty$ for some $L>0$. Then $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty b_n$ converges.

Proof

By definition there exists some $N\in\mathbb{Z}^+$ such that for $n > N$ we have for any $\varepsilon > 0$,

$\left|\frac{a_n}{b_n} - L\right| <\varepsilon.$

For simplicity, choose $\varepsilon =\frac{L}{2}$. We now have the inequality

$\left|\frac{a_n}{b_n} - L\right| <\frac{L}{2}.$

Unwrapping the modulus we get

$-\frac{L}{2} < \frac{a_n}{b_n} -L < \frac{L}{2}$

and adding $L$ yields

$\frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2}.$

As $b_n$ is strictly positive for every $n \in \mathbb{Z}^+$ we then have

$\frac{L}{2}b_n < a_n < \frac{3L}{2}b_n.$

Now, this holds for every $n > N$ so we deduce

$\frac{L}{2}\sum_{n=N+1}^{\infty}b_n < \sum_{n=N+1}^{\infty}a_n < \frac{3L}{2}\sum_{n=N+1}^{\infty}b_n.$

Suppose that $\sum_{n=1}^\infty a_n$ converges. Then we have

$\sum_{n=1}^{\infty}b_n =\sum_{n=1}^{N}b_n + \sum_{n=N+1}^{\infty}b_n$

and using the left hand side of the inequality and the fact that $a_n$ is always positive

$\sum_{n=1}^{\infty}b_n < \sum_{n=1}^{N}b_n + \frac{2}{L}\sum_{n=N+1}^{\infty}a_n$

$\sum_{n=1}^{\infty}b_n < \sum_{n=1}^{N}b_n + \frac{2}{L}\sum_{n=1}^{\infty}a_n.$

The two series on the right are convergent, and the sum of convergent series is convergent, so by the comparison test we conclude that $\sum_{n=1}^\infty b_n$ converges. Now suppose that $\sum_{n=1}^\infty b_n$ converges. We start with

$\sum_{n=1}^{\infty}a_n =\sum_{n=1}^{N}a_n + \sum_{n=N+1}^{\infty}a_n$

and overestimate, as by similar reasoning to before

$\sum_{n=N+1}^{\infty}a_n < \frac{3L}{2}\sum_{n=N+1}^{\infty}b_n <\frac{3L}{2}\sum_{n=1}^{\infty}b_n.$

We then deduce

$\sum_{n=1}^{\infty}a_n =\sum_{n=1}^{N}a_n + \frac{3L}{2}\sum_{n=1}^{\infty}b_n$

and again, the two series converge, and the sum of convergent series is convergent, so by the comparison test we conclude $\sum_{n=1}^\infty a_n$ converges.This completes the proof.