Find sum of $1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$ Find the sum of the following series : 
$$1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\cdots$$ and 
$$1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$$
For the first series , we have , $$(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}+\cdots$$
$$=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots=\frac{1}{2}\log 2$$
But I am unable to set the second series to find its sum..Please help.
 A: For conditionally convergent series, you can move terms around to a limited extent and still get the same sum.  Thus
$$1+{1\over3}+{1\over5}-{1\over2}-{1\over4}-{1\over6}+{1\over7}+{1\over9}+{1\over11}-\cdots=1-{1\over2}+{1\over3}-{1\over4}+{1\over5}-{1\over6}+{1\over7}-{1\over8}+\cdots$$
because no term is moving more than $3$ positions.  In this case the proof is also utterly straightforward:  The partial sum $s_n$ for each series is the same whenever $n$ is a multiple of $6$, so the series (each of which is convegent) must converge to the same sum.
A: The first is $\sum_{i=0}^{\infty}[\frac 1{2i+1} - \frac 1{2(2i+1)} - \frac 1{2(2i+1) + 2}]= \sum_{i=0}^{\infty}[\frac 1{2i+1} - \frac 1{2(2i+1)} - \frac 1{4(i+1)}]$
which is $\sum_{i=0}^{\infty}[\frac {4(i+1)}{4(2i+1)(i+1)} - \frac {2(i+1)}{4(2i+1)(i+1)} - \frac {2i+1}{4(2i+1)(i+1)}]=\sum_{i=0}^{\infty}\frac {1}{4(2i+1)(i+1)}$ which is ... I don't know.  Somehow you seem to believe $\sum \frac 1{2(2i+1)(i+1)} = \log 2$ which... I won't deny although I don't know how you got that.
The second is $\sum_{i=0}^{\infty}[\frac{1}{6i + 1} + \frac{1}{6i+3} + \frac{1}{6i+5} - \frac{1}{6i + 2} - \frac{1}{6i+4} - \frac{1}{6i + 6}] = \sum_{i=0}^{\infty}[\frac{1}{2(3i + 1) - 1} - \frac{1}{2(3i + 1)}$$ + \frac{1}{2(3i + 2) - 1} - \frac{1}{2(3i + 2)} + \frac{1}{2(3i + 3) - 1} - \frac{1}{2(3i + 3)}]$
$=\sum_{j= 1}^{\infty}(\frac{1}{2j-1} - \frac{1}{2j}) = \sum_{j=1}^{\infty}\frac 1{2j(2j-1)} = \frac 1 2 \sum_{i=0}^{\infty}\frac 1{(i+1)(2i+1)}$  which is... I don't know.  But again, I won't deny it is $\log 2$ although I don't know how you got that.
But the point to note, the second series is exactly twice the first series.
