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!
 A: $$\frac1{n(n+1)}=\frac1n-\frac1{n+1}\implies$$
$$\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)}=\sum_{n=1}^\infty\frac{(-1)^n}n-\sum_{n=1}^\infty\frac{(-1)^n}{n+1}=-\log2-(\log2-1))=1-2\log2$$
A: $$\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)} = \sum_{n=1}^\infty(-1)^n\left(\frac{1}{n} - \frac{1}{n+1}\right)$$
This is true because:
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
$$\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)} = \sum_{n=1}^\infty-\frac{(-1)^{n+1}}{n} - \frac{(-1)^n}{n+1} $$
First term converges to $- \log{2}$. Now we work on the second term.
If we set $n=k-1$, we can write:
$$\sum_{n=1}^\infty\frac{(-1)^n}{n+1} = \sum_{k=2}^\infty \frac{(-1)^{k+1}}{k}$$
Now we have:
$$\sum_{k=2}^\infty \frac{(-1)^{k+1}}{k} = \sum_{k=2}^\infty \frac{(-1)^{k+1}}{k} + \frac{1}{1} - \frac{1}{1} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} - \frac{1}{1} = \log 2 - 1$$
Now we substitute:
$$\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)} = - \log{2} - \log{2} + 1 = 1 - 2\log{2} = 1-\log{4}$$
A: $
\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}
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