# Verification of my proof of the Limit Comparison Test

The Limit Comparison Test: Let $$\{a_n\}$$ and $$\{b_n\}$$ be positive sequences, and suppose that $$\lim_{n\to\infty} \frac{a_n}{b_n} = c \neq 0$$ for some constant $$c$$, then $$\sum a_n$$ and $$\sum b_n$$ either both converge or both diverge.

First note that if $$a_n/b_n \to c$$, then that implies that for every $$\varepsilon > 0$$, there exists an $$N \in \mathbb{N}$$ such that if $$n \in \mathbb{N}$$ satisfies $$n \ge N$$, then $$|a_n/b_n - c| < \varepsilon.$$ This has to be true for every $$\varepsilon > 0$$, so it must be true for $$\varepsilon = c$$ as well (as $$a_n/b_n$$ is a ratio of two positive numbers).

Therefore, we have that $$\frac{a_n}{b_n} - c \le \left| \frac{a_n}{b_n} - c \right| < c,$$ where we applied the triangle inequality. So we have that $$\frac{a_n}{b_n} \le 2c \implies a_n \le 2cb_n. \quad\text{(for large enough }n)$$ By the Series Comparison Test, this tells us that if $$\sum b_n$$ converges, then $$\sum a_n$$ converges as well. Note that we also have that $$c - \frac{a_n}{b_n} \le \left| c - \frac{a_n}{b_n} \right| < c/2,$$ where we applied the triangle inequality. Using similar algebra as before, we get that $$b_n \le \tfrac{2}{c} a_n$$. By the Series Comparison Test, if $$\sum a_n$$ converges, then $$\sum b_n$$ will converge as well.

I do not have a teacher/professor, and I am just self-studying so I am looking for feedback on the proof above. Thank you, and feel free to be as pedantic as you would like.

## 1 Answer

Your idea seems fine but I don't see clearly stated the proof for the divergent case.

More simply, by definition of limit we have that eventually for $$n\ge n_0$$ we have $$a_n \le 2c\, b_n$$ and $$a_n \ge \frac c 2b_n$$ therefore by direct comparison test (i.e. squeeze theorem)

$$\sum_{n\ge n_0} b_n=\infty \implies \sum_{n\ge n_0} a_n\ge \frac c 2\sum_{n\ge n_0} b_n=\infty$$

$$\sum_{n\ge n_0} b_n=L \implies \sum_{n\ge n_0} a_n\le 2c\sum_{n\ge n_0} b_n=2cL$$

Edit

Proceeding in a similar way we can simply extend the result for the extreme cases

• $$\lim_{n\to\infty} \frac{a_n}{b_n} =0$$ $$\sum b_n<\infty \implies \sum a_n<\infty$$

• $$\lim_{n\to\infty} \frac{a_n}{b_n} =\infty$$ $$\sum b_n=\infty \implies \sum a_n=\infty$$

• Yes, you're right I should have explicitly written it out. Thanks for confirming. Commented Oct 6, 2021 at 14:11
• @politeproofs You are welcome! I add something else.
– user
Commented Oct 6, 2021 at 14:13