<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] When $|x| < 1$:
$$\begin{align*}\sum_{n=1}^\infty \dfrac{x^{n+1}}{n(n+1)} &= \sum_{n=1}^\infty \dfrac{x^{n+1}}{n} - \dfrac{x^{n+1}}{n+1} \\ &= x\sum_{n=1}^\infty \dfrac{x^n}{n} - \sum_{n=1}^\infty \dfrac{x^{n+1}}{n+1} \\ &= x\sum_{n=1}^\infty \dfrac{x^n}{n} - \sum_{n=1}^\infty \dfrac{x^n}{n} + x \\ & =  -x\ln(x-1) + \ln(x-1) + x\\ &= x+(1-x)\ln (1-x)\end{align*}$$
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