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Prove that: $$\sum\limits_{n=1}^\infty \dfrac{n^2}{2^n}=6.$$ I am trying to find the sum of this infinite series. Got stuck. Any help will be highly appreciated.

Martin Sleziak
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Mr. MBB
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    You can write $$\sum_{n=1}^\infty n^2x^n$$ in terms of the Taylor series of $\dfrac1{1-x}$ and its first and second derivative. –  Jul 12 '16 at 15:50

5 Answers5

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We have, $$S = \sum_{n=1}^\infty \frac{n^2}{2^n} = \sum_{n=0}^\infty \frac{n^2}{2^n} $$
$$S = \sum_{n=1}^\infty \frac{n^2}{2^n} = \sum_{n=0}^\infty \frac{(n+1)^2}{2^{n+1}} $$

Since $S = 2S - S $,
\begin{align} S&=\sum_{n=0}^\infty \frac{(n+1)^2}{2^n} - \sum_{n=0}^\infty \frac{n^2}{2^n} \\ &=\sum_{n=0}^\infty \frac{2n+1}{2^n}\\ &= \sum_{n=0}^\infty \frac{2n}{2^n} + \sum_{n=0}^\infty \frac{1}{2^n} \\&= 4+2 = 6 \end{align}

Mc Cheng
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    This is by for the best solution with a very clever trick Thanks, – Mr. MBB Jul 12 '16 at 16:08
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    I am very happy to see that you understand my answer :) – Mc Cheng Jul 12 '16 at 16:10
  • @Mr.MBB I was looking for the solution to this series just today!! Thank you so much for this elegant answer :) – P. N. Karthik Jul 12 '16 at 16:18
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    It is important to point out that doing this manipulation here is justified because the series is absolutely convergent; in contrast to the case where $S=1+2+3+\ldots$, where it is not. – Asaf Karagila Jul 12 '16 at 16:39
  • @Mr. Cheng I like the trick to reduce the degree of the polynomial in the numerator. – Mr. MBB Jul 12 '16 at 16:45
  • Also noted (though quite obvious): $2\sum\limits_{n=0}^{\infty} \frac{n}{2^n} = 4$ can be achieved the same way. $\sum\limits_{n=0}^{\infty} \frac{1}{2^n} = 2$ should be common knowledge. – Imago Jul 12 '16 at 17:01
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Hint: For $|x|<1$:

$\frac{1}{1-x} = \sum_{i=0}^\infty x^i$

$\frac{d}{dx} [\frac{1}{1-x}] = \sum_{i=1}^\infty i x^{i-1}$

$\frac{d^2}{dx^2} [\frac{1}{1-x}] = \sum_{i=2}^\infty i (i-1) x^{i-2}$

Take $x=\frac{1}{2}$ and use these statements with a change of indexing to get $\sum_{i=0}^\infty \frac{i^2}{2^i}$.

Batman
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Just for variety, reindexing sums and using the geometric series formula:

$$\begin{align} \sum_{n=1}^{\infty}\frac{n^2}{2^n} &=\sum_{n=1}^{\infty}\frac{\sum_{k=1}^{n}(2k-1)}{2^n}\\ &=\sum_{k=1}^{\infty}(2k-1)\sum_{n=k}^{\infty}\frac{1}{2^n}\\ &=\sum_{k=1}^{\infty}(2k-1)\frac{(1/2)^k}{1-1/2}\\ &=2\sum_{k=1}^{\infty}\frac{(2k-1)}{2^k}\\ &=2\sum_{k=1}^{\infty}\frac{k}{2^{k-1}}-2\sum_{k=1}^{\infty}\frac{1}{2^k}\\ &=2\sum_{k=1}^{\infty}\frac{\sum_{j=1}^k1}{2^{k-1}}-2\frac{1/2}{1-1/2}\\ &=2\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}\frac{1}{2^{k-1}}-2\\ &=2\sum_{j=1}^{\infty}\frac{1/2^{j-1}}{1-1/2}-2\\ &=4\sum_{j=1}^{\infty}\frac{1}{2^{j-1}}-2\\ &=4\frac{1}{1-1/2}-2\\ &=6 \end{align}$$

2'5 9'2
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prove by induction that $$\sum_{n=1}^k\frac{n^2}{2^n}=2^{-k} \left(-k^2-4 k+3\ 2^{k+1}-6\right)$$

Dr. Sonnhard Graubner
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Let $S=\sum_{n\geq 1}\frac{n^2}{2^n}$ (the series is obviously convergent). We have:

$$\begin{eqnarray*} S = 2S-S &=& \sum_{n\geq 1}\frac{n^2}{2^{n-1}}-\sum_{n\geq 1}\frac{n^2}{2^n}\\&=&1+\sum_{n\geq 1}\frac{(n+1)^2-n^2}{2^n}\\&=&1+\sum_{n\geq 1}\frac{2n+1}{2^n}=2+2\sum_{n\geq 1}\frac{n}{2^n},\end{eqnarray*}$$ hence it is enough to show that $T=\sum_{n\geq 1}\frac{n}{2^n}=2$. In the same way: $$ T = 2T-T = 1+\sum_{n\geq 1}\frac{1}{2^n} = 2 $$ and we are done. A similar proof is the following one: for any $x$ with $|x|<1$, $$ (1-x)^3\sum_{n\geq 1}n^2 x^n =\sum_{n\geq 1}\delta^3(n^2)\,x^n = x+x^2$$ with $\delta$ being the backward difference operator. That gives: $$ \sum_{n\geq 1}n^2 x^n = \frac{x+x^2}{(1-x)^3}$$ and it is enough to evaluate the previous identity at $x=\frac{1}{2}$.

Jack D'Aurizio
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