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?  
 A: I suppose you wanted to write that 
$1)$
if $\overline \lim(\frac {a_n}{b_n})<+\infty$ and $\sum b_n<+\infty$ then $\sum a_n$ converges too.
$2)$$\underline \lim(\frac {a_n}{b_n})>0$ and $\sum b_n$ diverges then $\sum a_n$ diverges too.
A: If $a_n,b_n\ge 0$ and $\lim_{n\rightarrow\infty} \frac{a_n}{b_n}=1$ then $\exists N_1$ such that, 
$\frac{a_n}{b_n}>\frac{1}{2}$ for $n\ge N_1$ 
which is equivalent to $\quad$$2a_n>b_n$, for $n\ge N_1$
hence, if $\sum_{n}^{\infty} a_n$ converges, then $\sum_{n}^{\infty} 2a_n>\sum_{n}^{\infty} b_n$ also converges.
Similarly $\exists N_2$ such that, $\frac{a_n}{b_n}<\frac{3}{2}$ for $n\ge N_2$
So $a_n<\frac{3b_n}{2}$
If $\sum_{n}^{\infty} b_n$ converges, then also $\sum_{n}^{\infty} a_n$
A: 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.
