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 
$$\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+1)(n+2)}=\left( 1-x \right)\log  \left( 1-x \right)  +x$$
$$\sum_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}=\left( 1-x \right)\log  \left( 1-x \right)  +x$$
$$\sum_{n=1}^{\infty}\frac{x^{n}}{n(n+1)}=\frac{\left( 1-x \right)\log  \left( 1-x \right)}x  +1$$
(Sum should be from $n=1$.)

$$\frac1{n(n+1)} = \frac1{n}-\frac1{n+1} $$

so that 

$$\begin{align}\sum_{n=1}^{\infty} \frac{x^n}{n (n+1)} &amp;= \sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1} \\ &amp;= -\log{(1-x)} - \frac1{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}\\ &amp;=-\log{(1-x)} - \frac1{x} \left [-\log{(1-x)}-x \right ]\\ &amp;= 1+\left ( \frac1{x}-1\right )\log{(1-x)} \end{align}$$

Note that the radius of convergence is $1$, i.e., the sum only converges when $|x| \le 1$.