Identify the function represented by $\displaystyle \sum_{k=2}^\infty \frac{x^k}{k(k-1)}$ So first I wrote it out in the terms, and I got
$\displaystyle \sum_{k=2}^\infty \frac{x^k}{k(k-1)} = \frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{20}+...$
I know the power series for $\displaystyle ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$ which is similar to the derivative of the power series from above, as
$\displaystyle \frac{d}{dx}(\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{20}+...)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...$
My question is, where do I go from here? How would I make it so the series is similar ln(1+x)? Or am I even going down the right route when it comes to solving this problem? Any help would be appreciated.

What you could have done is to differentiate twice to get
$$\left( \sum_{k=2}^\infty \frac{x^k}{k(k-1)}\right)''=\sum_{k=2}^\infty x^{k-2}=\frac 1 {1-x}$$making
$$\left( \sum_{k=2}^\infty \frac{x^k}{k(k-1)}\right)'=-\log(1-x)$$
$$ \sum_{k=2}^\infty \frac{x^k}{k(k-1)}=x+(1-x) \log (1-x)$$
