Proving $\lim_{n\to\infty}(\frac {a_n}{b_n}) \ne 0 \implies (\sum_{n=1}^\infty a_n =S_1 \iff \sum_{n=1}^\infty b_n =S_2) $

Question

I am having trouble proving the statement below for sequences $a_n\in \mathbb R,b_n\in \mathbb R^+ $. Assuming the following limit exists then :

$$x=\lim_{n\to\infty}(\frac {a_n}{b_n}) \ne 0 \implies (\sum_{n=1}^\infty a_n =S_1 \iff \sum_{n=1}^\infty b_n =S_2) $$

Whereas $S_1,S_2 $ should simply imply that each of the series converge.

I started out by assuming that $x\ne 0$ and $\sum_{n=1}^\infty a_n$ converges. Now what came to my mind was, since the series converges it implies that $\lim_{n\to\infty}(a_n) = 0$ . Wouldn't that lead to a contradiction at this point though, since I assumed that $x\ne 0$ and if $a_n$ converges to $0$ then it must follow that $x=0$ ,right? Any hints?

Answer 1

There is $N \in \mathbb N$ such that

$\frac{1}{2}|x| \le \frac{|a_n|}{|b_n|} \le \frac{3}{2}|x|$ for $n>N$. Thus

$\frac{1}{2}|x||b_n| \le |a_n| \le \frac{3}{2}|x||b_n|$ for $n>N$.

It follows:

$\sum_{n \ge 1}a_n$ converges iff $\sum_{n \ge 1}b_n$ converges

Answer 2

Hint

Take $\epsilon=\frac{x}{2}$

• use the definition of the limit.

• use comparison test for positive series.