calculus - 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.

$$\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1$$
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?

Answer

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.