$\displaystyle \sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n}$

I can figure out how to prove this series converges, but I cannot figure out how to find its sum.

I know this is the answer:

$\displaystyle \sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n} = \frac{log(2)}{2}$

But how?

I got a hint to use this: $\displaystyle tan^nu$ $\displaystyle du = \frac{1}{n-1}tan^{n-1}u-\int{tan^{n-2}}du$

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Proof it converges:

$\displaystyle b_n =\frac{1}{2n}$

$\displaystyle b_{n+1}=\frac{-1}{2n^2}$

Since $\displaystyle n=1$, $\displaystyle n$ is always positive,

$\displaystyle b_n > b_{n+1}$

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A hint to get me started would be awesome.