Splitting infinite sums When am I allowed to do this:
$$\sum_{n=1}^{\infty} f_n + g_n = \sum_{n=1}^{\infty} f_n + \sum_{n=1}^{\infty} g_n $$?
I know I can do it if both $\sum_{n=1}^{\infty} f_n$ and $\sum_{n=1}^{\infty} g_n$ are convergent, but can I do it if one of them is divergent and the other convergent?
 A: If one is convergent and one divergent the sum will diverge.  The place you have trouble is that both the sum of $f_n$ and the sum of $g_n$ can be divergent while the sum is nicely convergent.  For example $$f_n=2^n\\g_n=-2^n\\f_n+g_n=0$$
A: If any 2 sums exist, then the third exists and the equality holds. e.g. suppose $f_n+g_n$ and $f_n$ is summable. Then $-f_n$ is summable, and apply the known result for $F_n := f_n + g_n$ and $G_n = -f_n$ to see that $F_n + G_n$ is summable, with the equality.
A: It is even worse. Consider two divergent series: $\sum_{n} f_n = \sum_{n} \frac1n$ and  $\sum_{n} g_n = - M \sum_{n} \frac1{M n}$ with some natural number $M$. They both diverge, since what we have here are both harmonic series. Now add the two. What you get is 
$$
\sum_{n} f_n + g_n =  \sum_{n} \frac1n - M \sum_{n}\frac1{M n} \\
= 1 + \frac12 + \cdots + \frac{1}{M-1}+  \frac{1-M}{M} + \frac{1}{M+1} + \cdots + \frac{1}{2M-1}+  \frac{1-M}{2M} + \frac{1}{2M+1} + \cdots \\
= \ln(M) 
$$
So $\sum_{n} f_n + g_n $ is convergent. The last result may be surprising (had you expected it is zero?) but can be found here. A prominent special case is $M=2$ which is 
$$
\sum_{n} f_n + g_n =  \sum_{n} \frac1n - 2 \sum_{n}\frac1{2 n} \\
= 1 - \frac12 + \frac13 - \frac14 + \cdots = \sum_{n} \frac{(-1)^{n}}{n}
= \ln(2) 
$$
which is the well known alternating harmonic series. 
