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!

$
\begin{array}{l}
 \frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}} \\ 
  \Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{{n\left( {n + 1} \right)}}}  = \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}}  - \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{{n + 1}}}  \\ 
  \Rightarrow  - \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}}  = \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}}  + \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 2} }}{{n + 1}}}  \\ 
  \Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}}  =  - \left( {\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}}  + \sum\limits_{n = 2}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} } \right) \\ 
  \Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}}  =  - \left( {\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}}  - 1 + \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} } \right) \\ 
  \Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}}  =  - \left( {2\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}}  - 1} \right) \Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}}  = 1 - 2\log 2 = 1 - \log 4 \\ 
 \end{array}
$
