Sum of a sequence I need guidance for the following question.

Using the fact that $\sum_1^{\infty}\frac{(-1)^{n+1}}{n}=\log2$, $\sum_1^{\infty}\frac{(-1)^{n}}{n(n+1)}$ equals
$1.$ $1-2\log2$
$2.$ $1+2\log2$
$3.$ $(\log2)^2$
$4.$ $-(\log2)^2$

The given sequence gives us $1-\frac12+\frac13-\frac14+\cdots=log2$, but I am unable to think how this would help me to solve $-\frac12+\frac16-\frac1{12}+\frac1{20}-\cdots$
I wish somebody could help. Thanks in advance!

$$\frac1{n(n+1)}=\frac1n-\frac1{n+1}\implies$$
$$\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)}=\sum_{n=1}^\infty\frac{(-1)^n}n-\sum_{n=1}^\infty\frac{(-1)^n}{n+1}=-\log2-(\log2-1))=1-2\log2$$
