Infinite sum of $\frac{x^n}{n+1}$

Question

I was solving a calculus problem and found myself stuck trying to find to solve:

$$S = \sum_{n=0}^{\infty} \frac{x^n}{n+1}$$

Wolfram Alpha says:

But I have no idea how this sum is evaluated. Any help is appreciated.

Answer 1

$\frac d {dx} (xS)=\frac d {dx} \sum \frac {x^{n+1}} {n+1}=\sum x^{n}=\frac1 {1-x}$. Hence $xS=\int \frac1 {1-x} \, dx=-\ln (1-x)+C$. This gives $S=-\frac {\ln(1-x)} x+\frac C x$. Note that $S=1$ when $x=0$, so $C=0$.

Answer 2

Reverse engineering the answer:

The answer contains a logarithmic function, which is transcendental. We can try to let it disappear by differentiation.

Anyway,

$$S'(x)=-\left(\frac{\log(1-x)}x\right)'=\frac{1}{x(1-x)}+\frac{\log(1-x)}{x^2}$$ doesn't have the intended effect.

Now,

$$(xS(x))'=-(\log(1-x))'=\frac1{1-x}$$ is more interesting. Let us compare with the term-wise differentiation of the series,

$$\left(x\sum_{n=0}^\infty\frac{x^n}{n+1}\right)'=\left(\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}\right)'=\sum_{n=0}^\infty x^n$$ and we recognize a geometric series.