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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Which one of the above results do you mean? – Mikasa Jan 08 '14 at 12:16
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Apparently, there was some unfortunate editing here. http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427etrd0bedh67 – user3071205 Jan 08 '14 at 12:19
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Did you mean this? – Michael Albanese Jan 08 '14 at 12:20
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@user3071205: But this link is not the same as you added in the body. :-) – Mikasa Jan 08 '14 at 12:21
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Does my edit ask the question you wanted to ask? – Michael Albanese Jan 08 '14 at 12:23
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yes. I added the correct link in the original post. – user3071205 Jan 08 '14 at 12:23
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If $f(z) = \sum_{k=0}^{\infty} a_k z^k$, then $z\frac{df(z)}{dz} = \sum_{k=0}^{\infty} k a_k z^k$. Since we know what $\sum_{k=0}^{\infty} z^k$ is..... – achille hui Jan 08 '14 at 12:32
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1I rolled back the edit because it is better to have the formula here than to have a link to another site which displays it. – Michael Albanese Jan 08 '14 at 12:35
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@MichaelAlbanese: Thanks for the edit. – Mikasa Jan 08 '14 at 12:56
2 Answers
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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}
Tim Ratigan
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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.
Mhenni Benghorbal
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