I would like to stress one additional thing which is not included in the above answers.
Assuming that $\sum_{i=1}^\infty a_i$ and $\sum_{i=1}^\infty b_i$ are convergent to say $A$ and $B$, we know that $\sum_{i=1}^\infty (a_i + b_i)$ is convergent and its limit is $A+B$. So if we only know that the problem of finding $\sum_{i=1}^\infty a_i + \sum_{i=1}^\infty b_i$ is well defined (series are convergent), then we have: $\sum_{i=1}^\infty a_i + \sum_{i=1}^\infty b_i = \sum_{i=1}^{\infty} (a_i+b_i).$
A little bit more is also true: if $\sum a_i$ and $\sum b_i$ diverges to infinity or converges to some value AND the sum $A+B$ can be defined (ie. it is not $\infty - \infty$ or sth similar. $-\infty - \infty=-\infty$, $-5 + \infty = \infty$ etc. are OK), then the above is also true.
EDIT
Since I was downvoted, I think I need to give a proof.
First for convergent series. $A$ is said to be the value of $\sum_{i=1}^\infty a_i$ if $\lim_{n \to \infty} \sum_{i=1}^n a_i = A$. Let $\varepsilon>0$. From the assumption we have $N$ satisfying: for $n>N$ $|\sum_{i=1}^n a_i - A| < \frac{\varepsilon}{2}$ and the same for $\sum b_i$. So we have that for $n>N$: $|\sum_{i=1}^n (a_i+b_i) - (A+B)| = |\sum_{i=1}^n a_i + \sum_{i=1}^n b_i - (A+B)| =$ $ = |\sum_{i=1}^n a_i - A + \sum_{i=1}^n b_i - B| \leq |\sum_{i=1}^n a_i - A| + |\sum_{i=1}^n b_i - B| \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$
Second - if one of them (say $\sum a_i$) diverges to infinity and the other diverges to infinity or converges to a finite value. (The case with minus infinity/infinities is the same.) Take any $M < \infty$. If $\sum_{i=1}^\infty b_i$ is $\infty$, then $k=0$. Otherwise $k=\sum_{i=1}^\infty b_i$. Let $P=M-k+1$.
From the assumptions we know that there exists $N$ such that for $n>N$ we have: $\sum_{i=1}^n a_i > P$ and $\sum_{i=1}^n b_i > k-1$. Consequently $\sum_{i=1}^n (a_i+b_i) = \sum_{i=1}^n a_i + \sum_{i=1}^n b_i > P + k-1 = M$, qed.