Does $a_n \sim b_n$ imply $\sum_n a_n \sim \sum_n b_n$ for $a_n, b_n>0$? I am almost embarrassed to asked to this question, but after considering it for a while I realize I need some help.
In the following $a_n, b_n >0$.
So, by limit comparison test if $a_n = O(b(n))$ or, for a stricter case that I am interested in, $\frac{a_n}{b_n} \to_n 1$ both $S_1 = \sum_n a_n$ and $S_2 = \sum_n b_n$ either diverge or converge or, in other words, $S_1 = O(S_2)$ or $\frac{S_1}{S_2} \to_n O(1)$. 
How about a stricter condition: does this convergence imply that $\frac{S_1}{S_2} \to_n 1$ or $S_1 \sim S_2$?. For example, $S_1 \sim n \log n + O(1)$ and $S_2 \sim n \log n  + n + \frac{1}{n}$. 
I think the answer is not in general, I tried to compare $\sum_k \log (1+\frac{k}{n})$ and $\sum_k \frac{k}{n}$ (they are $\sim$ if Taylor series is used). And I got the first constant to be $\log \frac{4}{e}$, and the second, of course, $\frac{1}{2}$. 
Any suggestions? 
EDIT:
After reading the solution, I went through my calculation again and what I get is 
$$
\sum_{k=1}^{n} \log (1+\frac{k}{n}) = \log (2n)! - \log n! - n \log n 
$$
Either using Stirling's approximation or $\sum_{k=1}^{n} \log k  \sim \int_{1}^{n} \log x dx$ I get the front term equal to $\log \frac{4}{e}$, which is different to $\frac{1}{2}$ from $\sum_{k=1}^{n} \frac{k}{n}$. Where is the mistake? 
 A: This was the answer before OP had positivity conditions for $a_n$, $b_n$.
Recall that the limit comparison test usually applied when the terms are positive. So, by making some negative terms, we might easily get a counterexample. Consider this, 
$$
a_n = (-1)^n + \frac1n
$$
and
$$
b_n= (-1)^n
$$
Then $a_n/b_n = 1+(-1)^n / n \rightarrow 1$.  However,
$$
\sum_{n\leq N} a_n = \frac{1-(-1)^N}2 + \sum_{n\leq N} \frac1n = \log N + O(1)$$
and
$$
\sum_{n\leq N} b_n = \frac{1-(-1)^N}2 = O(1).$$
Answer after the positivity conditions.
Let $S_{1,N}=\sum_{n\leq N} a_n$ and $S_{2,N}=\sum_{n\leq N} b_n$. 
If $a_n = O(b_n)$, then we have for some positive constant $C$, 
$$
a_n \leq C b_n.$$
Thus, 
$$
\sum_{n\leq N} a_n \leq \sum_{n\leq N} C b_n, $$ 
which is $S_{1,N}=O(S_{2,N})$. 
Similarly if $a_n\sim b_n$, then by limit comparison test, 
$\sum a_n$ and $\sum b_n$ both converge, or both diverge. 
Since the case both converge is not of our interest: (both being $O(1)$.)
We consider the case both diverge. 
For any $\varepsilon>0$, there is $N_0$ such that if $n\geq N_0$, then
$$ 1-\varepsilon \leq \frac {a_n}{b_n} \leq 1+\varepsilon.$$
Let $N> N_0$. We have by the above inequalities, 
$$
(1-\varepsilon)\sum_{N_0< n\leq N} b_n \leq \sum_{N_0< n\leq N} a_n \leq (1+\varepsilon)\sum_{N_0< n \leq N} b_n.$$
Then we have 
$$
(1-\varepsilon)(S_{2,N}-S_{2,N_0}) \leq S_{1,N}-S_{1,N_0}\leq (1+\varepsilon)(S_{2,N}-S_{2,N_0}).
$$
Dividing everything by $S_{2,N}$, we have 
$$
(1-\varepsilon)\left( 1-\frac{ S_{2, N_0} }{S_{2,N}} \right)\leq \frac{S_{1,N}-S_{1,N_0}}{S_{2,N}}\leq (1+\varepsilon)\left( 1-\frac{ S_{2, N_0} }{S_{2,N}} \right)
$$
Letting $N\rightarrow\infty$, we have
$$
1-\varepsilon \leq \liminf_{N\rightarrow\infty} \frac{S_{1,N}}{S_{2,N}} \leq \limsup_{N\rightarrow\infty} \frac{S_{1,N}}{S_{2,N}} \leq 1+ \varepsilon.
$$
Since $\varepsilon>0$ is arbitrary, we obtain that
$$
\lim_{N\rightarrow\infty} \frac{S_{1,N}}{S_{2,N}} = 1.
$$
Therefore, in this case, $S_{1,N} \sim S_{2,N}$. 
A: I don't think it holds. 
Take $ a_{n} = \frac{1}{n!} + \frac{1}{(n!)^{2}} $ and $ b_{n} = \frac{1}{n!}  $.
Therefore, 
$$ \lim_{n\to\infty} \frac{a_{n}}{b_{n}} = 1 $$
and
$$ S_{a}(N)=\sum_{n=0}^{N}a_{n} \Rightarrow \lim_{N\to\infty}S_{a}(N) = e+\alpha $$
$$ S_{b}(N)=\sum_{n=0}^{N} b_{n} \Rightarrow \lim_{N\to\infty}S_{a}(N) = e $$
However, 
$$ \lim_{n\to\infty} \frac{S_{a}(n)} {S_{b}(n)} = 1+\frac{\alpha}{e} 
 $$
