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 $1 - 2 \log 2$.