Verification of my proof of the Limit Comparison Test

Question

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.

Answer 1

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

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

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