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.

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.
