<s>[INST] Calculate the sum of $\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$ I would like to calculate the sum for $x\in(-1,1)$ of this:  $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$$
So far I managed that 
$$\int \frac{x^n}{n}dx = \frac{x^{n+1}}{n(n+1)}, $$
and
$$\sum_{n=1}^\infty \frac{x^n}{n}=-\log(1-x), $$
and
$$ \int -\log(1-x) dx \overset{c}{=} x+(1-x)\log(1-x). $$
But I really don't know if I can put it somehow together or not.
 [/INST] Using partial fraction decomposition $$\frac 1{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ So
$$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty} \frac{x^{n+1}}n-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=x\sum_{n=1}^{\infty} \frac{x^{n}}n-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=x \log(1-x)-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$$ Now, $$\frac d{dx} \left(\frac{x^{n+1}}{n+1}\right)=x^n\implies \frac d{dx} \left(\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} \right) =\frac{x}{1-x}=\frac{x-1+1}{1-x}=\frac{1}{1-x}-1$$ Now, integrate to get $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=-\log(1-x)-x$$
</s>