# Series convergence properties of sequences whose ratio converges

Suppose $$(a_n)$$ and $$(b_n)$$ are strictly positive real-valued sequences and the ratio of these sequences converges, i.e.,

$$0 < \lim \frac{a_n}{b_n} < \infty$$

I'm trying to prove that therefore the series $$\Sigma\,a_n$$ and $$\Sigma\,b_n$$ either both converge or both diverge.

So far, I'm hitting a wall when trying to deduce anything about the sequences by assuming things about the series.

For instance, if I assume $$\Sigma\,a_n$$ converges but $$\Sigma\,b_n$$ diverges, I deduce only that $$lim\,a_n = 0$$ and $$b_n > a_n$$ by the contrapositive of the comparison test (and vice-versa if I assume the opposite).

The other approach I'm considering is to start with $$\frac{a_n}{b_n}$$ being Cauchy, but I don't know how to say anything about the individual series from there, either.

Any hints or recommendations as to what I'm missing would be greatly appreciated!

• How are you using that the limit of the ratio is finite in your thoughts so far? How can you express the fact that the ratio is finite in a may which will be useful to you? – Mark Bennet Nov 6 '18 at 7:30
• Yes, I see now that this is just a restatement of the limit comparison test. I knew of the direct comparison test, but hadn't seen an exact formulation of the limit one before. Thank you! – notadoctor Nov 6 '18 at 7:47

## 3 Answers

Assume that $$\lim_{n\to\infty}\frac{a_n}{b_n}=L>0$$. There exist some $$n_0\in\Bbb N$$ such that $$\frac L2<\frac{a_n}{b_n}<2L$$ for $$n\ge n_0$$.

Then, if $$\sum a_n$$ converges, $$\sum_{n=n_0}^\infty b_n<\frac2L\sum_{n=n_0}^\infty a_n$$ so $$\sum b_n$$ also converges.

Can you finish?

• there is a small typo for $n\ge n_0$ – user Nov 6 '18 at 7:46
• Yes, thank you! I somehow overlooked that this is just a restatement of the limit comparison test. I can finish the proof from here. – notadoctor Nov 6 '18 at 7:47

The hypothesis implies that there are positive constants $$c_1$$ and $$c_2$$ and an integer $$m$$ such that $$c_n \leq \frac {a_n} {b_n} \leq c_2$$ for $$n \geq m$$. For convergence of a series you can always omit the first few terms. So use comparison test by ignoring first $$m-1$$ terms. Hint for proving that $$c_1$$ and $$c_2$$ exist: let $$L=\lim \frac {a_n} {b_n}$$ and take $$c_1=L-\epsilon, c_2=L+\epsilon$$ where $$0<\epsilon . Show existence of $$m$$ using definition of limit.

As already shown we can prove easily the statement by the definition of limit (note that only $$b_n$$ is required to be strictly positive).

Note that as an extension also the following hold

1. $$\lim \frac{a_n}{b_n} =0$$ and $$b_n$$ converges then $$a_n$$ converges

indeed

$$\frac{a_n}{b_n}\le \epsilon \implies \sum a_n\le\epsilon \sum b_n$$

1. $$\lim \frac{a_n}{b_n} =\infty$$ and $$b_n$$ diverges then $$a_n$$ diverges

indeed

$$\frac{a_n}{b_n}\ge M \implies \sum a_n\ge M \sum b_n$$