Getting $f(x) =\sum_{n=2}^{\infty} \dfrac{x^n}{n(n-1)}$, $|x|<1$.

Question

In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.

What is $f(x)$ itself?

Here is the Taylor series at $x=0$ for the function:

$$ f(x)=\sum_{n=2}^{\infty} \dfrac{x^n}{n(n-1)} $$

I'm not sure about the $1$ in $n-1$.

Answer 1

Hint. One may recall that $$ \sum_{n=1}^\infty\frac{x^n}{n}=-\ln\left(1-x \right),\quad |x|<1. $$ Then writing $$ \frac{x^n}{n(n-1)}=x\frac{x^{n-1}}{n-1}-\frac{x^n}{n},\quad n\ge2, $$ may help.