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How do I calculate the infinite sum of this series?

$$\sum_{n=1}^{\infty}n^2q^n = -\frac{q(q+1)}{(q-1)^3}\ \text{when}\ |q| < 1.$$

How does wolfram alpha get this result?

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  • $\begingroup$ Which one of the above results do you mean? $\endgroup$ – mrs Jan 8 '14 at 12:16
  • $\begingroup$ Apparently, there was some unfortunate editing here. wolframalpha.com/share/… $\endgroup$ – user3071205 Jan 8 '14 at 12:19
  • $\begingroup$ Did you mean this? $\endgroup$ – Michael Albanese Jan 8 '14 at 12:20
  • $\begingroup$ @user3071205: But this link is not the same as you added in the body. :-) $\endgroup$ – mrs Jan 8 '14 at 12:21
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    $\begingroup$ I rolled back the edit because it is better to have the formula here than to have a link to another site which displays it. $\endgroup$ – Michael Albanese Jan 8 '14 at 12:35
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$$ \sum_{n=1}^\infty n^2q^n=q\sum_{n=1}^\infty n^2q^{n-1}=q\cdot\frac{\text d}{\text dq}\left(\sum_{n=1}^\infty nq^n\right) $$ It is well-known that $\frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$ (given $|x|<1$), so we have \begin{align} \sum_{n=1}^\infty n^2q^n&=q\cdot\frac{\text d}{\text dq}\left(\frac q{(1-q)^2}\right)\\ &=q\cdot\frac{(1-q)^2+2q(1-q)}{(1-q)^4}\\ &=\frac{q(1+q)}{(1-q)^3} \end{align}

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A related problem. Recalling the series

$$ \sum_{k=1}^{\infty} q^k = \frac{q}{1-q} $$

Now, apply the operator $(qD)^2 = (qD)(qD) $ where $D= \frac{d}{dq} $ to both sides of the above equation.

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