If $\lim\limits_{n \to \infty} \frac{a_n}{b_n}=1 \rightarrow \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n$ If $\lim\limits_{n \to \infty} \frac{a_n}{b_n}=1 \rightarrow \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n$
How may I prove this? Or give an example where it doesn't apply
Taking into account that $a_n > 0, b_n > 0, \forall n \in N$
Thanks in advance
 A: Take $a_n=\frac{1}{2^n}$ and $b_n=a_n$ for $n>1$ and $b_0=3$. Then $lim_{n\to \infty}\frac{a_n}{b_n}=1$ clearly and $\sum a_n=\sum b_n+2$.
A: This is false. Consider the sequences $a_n = \frac{1}{3^n} $ and $b_n = \frac{1}{3^n - n}$, then 
$$
\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1}{3^n} \frac{3^n - n}{1} = \lim_{n \to \infty} \left( 1 - \frac{n}{3^n} \right) = 1,
$$
but their sums are clearly different. Indeed, the first sum is a geometric series, so $\sum_{n=0}^{\infty} a_n = \frac{3}{2}$; the second sum can be calculated numerically to be $\sum_{n=0}^{\infty} b_n \approx 1.70378$.
A: this doesn't work in general, your series don't even have to exist, take
$a_n=b_n=2$ then $\sum_{n=1}^{\infty}a_n$ doesn't exists.
Or another example $a_n=b_n=2+(-1)^n$, the same problem with none existing sums...so you need at the very least that the sequences $a_n,b_n$ are converging to $0$ for having the series to be finite.
Okay, I just saw @Ian's comment, you could of course easily change finitely many $a_n$'s and  $b_n$'s. So it actually pretty much never works.
