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.
 A: Hint A. You have $f'(x)=-\ln(1-x)$. Can you integrate this?
Hint B. Alternatively, use partial fraction decomposition on $\displaystyle\frac{1}{k(k-1)}$.
A: 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
$$\left( \sum_{k=2}^\infty \frac{x^k}{k(k-1)}\right)'=-\log(1-x)$$
$$ \sum_{k=2}^\infty \frac{x^k}{k(k-1)}=x+(1-x) \log (1-x)$$
A: You're very close.  Since $$\log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots,$$ we have $$- \log (1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots.$$  Then integration term by term gives $$- \int \log(1-x) \, dx = \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{12} + \frac{x^5}{20} + \cdots.$$  So all you need to do is figure out how to perform the integration.  Do be careful, since the RHS should be zero when $x = 0$, so an antiderivative on the LHS should also be zero when $x = 0$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{k = 2}^{\infty}{x^{k} \over k\pars{k - 1}} & =
\sum_{k = 2}^{\infty}{x^{k} \over k - 1} -
\sum_{k = 2}^{\infty}{x^{k} \over k} =
\sum_{k = 1}^{\infty}{x^{k + 1} \over k} -
\pars{-x + \sum_{k = 1}^{\infty}{x^{k} \over k}}
\\[5mm] & =
\pars{1 - x}\pars{-\sum_{k = 1}^{\infty}{x^{k} \over k}} + x
\\[5mm] & =\
\bbox[10px,#ffd,border:1px groove navy]{\large\pars{1 - x}
\ln\pars{1 - x} + x} \\ &
\end{align}
