# What is the sum of the power series: $\sum_{k=2}^\infty\frac{x^k}{k(k-1)}$?

What is the function represented by the power series $$\sum_{k=2}^\infty\frac{x^k}{k(k-1)}\quad?$$

It looks like $\dfrac1{1-x}$ but I don't know.

One may write, as $$|x|<1$$, $$\sum_{k=2}^\infty\frac{x^k}{k(k-1)}=\sum_{k=2}^\infty\frac{x^k}{k-1}-\sum_{k=2}^\infty\frac{x^k}{k}=x\sum_{k=1}^\infty\frac{x^k}{k}-\sum_{k=2}^\infty\frac{x^k}{k}=(x-1)\sum_{k=1}^\infty\frac{x^k}{k}+x$$ and one may use
$$\sum_{k=1}^\infty\frac{x^k}{k}=-\log(1-x), \quad |x|<1$$
Set $$f(x)=\sum_{k=2}^\infty\frac{x^k}{k(k-1)}$$ Taking derivatives twice, we get $$f''(x)=\sum_{k=2}^\infty x^{k-2}=1+x+x^2+\cdots=\frac{1}{1-x}$$ Hence, to recover $f(x)$, we integrate $f''(x)=(1-x)^{-1}$ twice: $$f'(x)=-\ln (1-x)+C$$ $$f(x)= (C+1)x+(1-x)\ln(1-x)+D$$ We now need the constants. Taking $x=0$ we have $$0=f(0)=D$$ To find $C$ we look at $$f'(x)=\sum_{k\ge 2} \frac{x^{k-1}}{k-1}$$ We have $0=f'(0)=-\ln(1-0)+C$, so $C=0$ as well.