Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{n}}{n(n+1)}$ on its domain of convergence.
This is my idea.
We have the radius of convergence is $R=1$. And $\sum_{n=1}^{\infty} \dfrac{x^{n}}{n(n+1)}=\dfrac{1}{x}\cdot \sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.
Then let $f(x)=\sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.
We have $f'(x)=\sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ and $f''(x)=\sum_{n=1}^{\infty} x^{n-1}=\dfrac{1-x^n}{1-x}$
Then if I take the primity of $f''(x)$, I get: $f'(x)=\displaystyle\int\limits_{0}^{x}\dfrac{1-t^n}{1-t}\mathrm{d}t=-\ln(1-x)-\displaystyle\int\limits_{0}^{x}\dfrac{t^n}{1-t}\mathrm{d}t$
Then I will get a sum again, and that's not the goal.
What can I do then? Thank you so much.
If you know (or can assume) the logarithm series for $|x|<1$:
$$-\log(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$$
then you can forego the derivation/integration steps by noticing that:
$$ \begin{align} \sum_{n=1}^{\infty} \frac{x^{n}}{n(n+1)} & = \sum_{n=1}^{\infty} x^n \left(\frac{1}{n} - \frac{1}{n+1} \right) \\ & = \sum_{n=1}^{\infty} \frac{x^n}{n} - \frac{1}{x} \sum_{n=2}^{\infty} \frac{x^n}{n} \\ & = -\log(1-x) - \frac{1}{x}(-\log(1-x) - x) \\ & = \left(\frac{1}{x}-1\right)\log(1-x) \;+\; 1 \end{align} $$