Find sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{n(n+1)}$ 
I need to find sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{n(n+1)}$

I'm getting it as $2\ln2-1$

The sequence converges absolutely by comparison with $\sum_{n=1}^\infty \frac{1}{n^2}$. Note
$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
Therefore $$\sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} = \sum_{n=1}^\infty \left(\frac{1}{2n}-\frac{1}{2n-1}\right)+\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=1+2\sum_{n=1}^\infty\frac{(-1)^{n}}{n}=1-2\ln(2)$$
The second to last equality follows by noticing that all fractions of the forms $\frac{1}{2n}$ and $\frac{-1}{2n-1}$ appear twice, except $-1$, which is counted only once.
