TITLE: True/False: $\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1$ implies $\sum {{a_n},\sum {{b_n}} } $ converge or diverge together.
QUESTION [5 upvotes]: $$\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1$$
Prove the statement implies $\sum {{a_n},\sum {{b_n}} } $ converge or diverge together.
My guess the statement is true.  
if $\sum{{a_n}}$ diverges, then $\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0$  
So, 
$$\eqalign{
  & \mathop {\lim }\limits_{n \to \infty } {a_n} = L \ne 0  \cr 
  & {{\mathop {\lim }\limits_{n \to \infty } {a_n}} \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow {L \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {b_n} \ne 0 \cr} $$
therefore, 
$\sum {b_n}$ also diverges.  
What I was not managed to do is proving that the two series converges together.
Or maybe the statement is not always true?

REPLY [11 votes]: Surprisingly, this statement is false. For a simple counter-example, consider
$$
a_n = \frac{(-1)^n}{\sqrt{n}},\quad\text{and}\quad b_n = \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n}
$$
The condition $a_n \sim b_n$ holds but $\sum a_n$ is convergent whereas $\sum b_n$ is divergent.