Summation of alternating series, Mercator series: $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}$

Question

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.

Answer 1

Hint. Due to properties of power series, one is allowed to write $$ \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}&=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_{0}^1t^{n}dt \\\\&=\int_{0}^1\sum_{n=1}^\infty\frac{(-1)^{n+1}t^{n}}{n}\:dt \\\\&=\int_0^1\ln(1+t)dt \end{align} $$ Can you finish it?

Answer 2

It is well-known fact that

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

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