sum $1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + ...$ find the sum if the following series converges:
$$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + ...$$
my attempt:
The series can be grouped into difference of a series of odd terms and a series of even terms.. but how to find sum?

Let $L=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\dots \tag{1}$
$(1)$ converges by Alternating series test, so $L$ is finite
Go through this link to convince yourself that $L=\ln(2).$
Now multiply $L$ by $\frac{1}{2}$, we get $$\frac L2=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}\dots \tag 2$$
Now adding $(1)$ and $(2)$, we get 
$$\frac{3}2L=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}\dots \tag 3$$
Your question ends here as you have found the required sum as $\frac{3}2(\ln(2)) \approx 1.0397$

Now notice that $(3)$ is just a rearrangment of $(1)$, in a way, we list the first two +ve terms followed by first -ve term from $(1)$ to make up first three terms of $(3)$ , and then next two +ve terms followed by second -ve term, and so on.
Thus we see that rearrangments of a series are not necessarily to converge on the same limit. And thus, an interesting question pops up- "when does rearrangements converge to same limit?"
Answer- 

 When original series converges absolutely. 

