compute the sum of a series $$
\sum_{n=0}^\infty \frac{x^n}{n(n+1)}
$$
what is the sum of this?
I suspect this has something to do with $e^x = \sum x^n/n! $ but I don't know how to go from there

$$
\sum\limits_{n=0}^{\infty}x^n=\frac1{1-x}
$$
Integrate once to obtain
$$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log\left({1-x}\right)$$
Integrate again to obtain 
$$\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+1)(n+2)}=\left( 1-x \right)\log  \left( 1-x \right)  +x$$
$$\sum_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}=\left( 1-x \right)\log  \left( 1-x \right)  +x$$
$$\sum_{n=1}^{\infty}\frac{x^{n}}{n(n+1)}=\frac{\left( 1-x \right)\log  \left( 1-x \right)}x  +1$$
