Series convergence properties of sequences whose ratio converges

Question

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!

Answer 1

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?

Answer 2

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 $0<\epsilon <L$. Show existence of $m$ using definition of limit.

Answer 3

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

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

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

indeed

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