I have to use
$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}=\log 2$
to compute
$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n(n+1)}$
Since,
$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n(n+1)} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n+1)}$
Now, $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} = \ -log 2 $
$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n+1)} = \log 2 -1 $
Hence , the answer seems to be $1 - 2 \log 2$.