Summation of alternating series, Mercator series: $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}$ I am struggling with solving sum of this alternate series:
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}\ 
$$
I know that:
$$
\log(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \cdot x^n\
$$
But It seems that I can't find a way to get to this form.
Thanks.

It is well-known fact that

$${ \sum _{ n=1 }^{ \infty  }{ \frac { (-1)^{ n+1 } }{ n }  } =\ln { 2 } . }$$

So 
$$\sum _{ n=1 }^{ \infty  } \frac { (-1)^{ n+1 } }{ n(n+1) } =\sum _{ n=1 }^{ \infty  } \left( \frac { (-1)^{ n+1 } }{ n } -\frac { (-1)^{ n+1 } }{ n+1 }  \right) =\sum _{ n=1 }^{ \infty  } \frac { (-1)^{ n+1 } }{ n } -\sum _{ n=1 }^{ \infty  } \frac { (-1)^{ n+1 } }{ n+1 } =\left( 1-\frac { 1 }{ 2 } +\frac { 1 }{ 3 } -\frac { 1 }{ 4 } +.. \right) -\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } +\frac { 1 }{ 4 } +.. \right) =\\ =\ln { 2 } -\left( -\ln { 2 } +1 \right) =2\ln { 2 } -1\\ $$
