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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.

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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

$$ \sum_{k=1}^\infty\frac{x^k}{k}=-\log(1-x), \quad |x|<1 $$

to conclude.

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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.

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