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I understand that $\sum\limits_{i=0}^{\infty} x^i$ is $1/x-1$ and I know that $\sum\limits_{i=0}^{\infty} i\cdot x^i$ is $x/(x-1)^2$ but the $i^2$ in the original problem throws me off and I don't know how to differentiate to get it. Can someone show me the solution?

  • Please don't post the same question twice. If you want to clarify your previous question you can edit it by clicking on the "edit" link immediately below the list of tags below your question. – Antonio Vargas Sep 20 '15 at 02:29
  • Actually, $\sum x^i = \frac{1}{1-x}$ not $\frac{1}{x-1}$. – Thomas Andrews Sep 20 '15 at 02:29

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Hint $i^2x^i = x\cdot \left(\frac{d}{dx}(ix^i)\right)$.

Quang Hoang
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