Factor out the 2, first of all. Secondly, all the powers of -1 are odd, and hence equal to -1:
Then you have the sum (note k cannot equal zero... error?):
\(\displaystyle \frac{1}{2}\sum_{k>1}\frac{(-1)^{2k-1}}{k} = -\frac{1}{2}\left [ \frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +... \right ]\)
Look familiar now?
