How to go about summing $\sum_{k=0}^{\infty} \frac{k^2}{2^k}$.
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1What have you tried towards solving this question? – Parcly Taxel Jan 09 '20 at 16:54
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6Hint: Take two derivatives. – Qi Zhu Jan 09 '20 at 16:54
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This will get you started: https://math.stackexchange.com/a/337942/16192 – gt6989b Jan 09 '20 at 16:55
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See also https://math.stackexchange.com/questions/439062/summation-of-series-sum-n-1-infty-fracnabn – Arnaud D. Jan 09 '20 at 16:56
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Do you know any calculus? Are you familiar with $1+\frac{1}{2}+\frac{1}{2^2}+\dots=2$? – almagest Jan 09 '20 at 16:58
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Yes. I'm aware of the above relation. – sbp Jan 09 '20 at 17:01
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Thanks for the hint. $x^2 f^{''}(x)$ gives me the summation that I need. But the index now starts from $2$. Any tip regarding that? – sbp Jan 09 '20 at 17:49
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My bad! Got it, thanks! Answer will be $6$. – sbp Jan 09 '20 at 18:00
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You're going to have to do summation by parts here. – Allawonder Jan 09 '20 at 18:16
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https://math.stackexchange.com/questions/1500335/sum-up-the-following-series-sumn-k-1-frack22k and https://math.stackexchange.com/questions/1072038/infinite-series-sum-k-1-infty-frack2k-and-sum-k-1-infty-frack and https://math.stackexchange.com/questions/2243801/how-to-determine-the-sum-of-an-infinite-series-that-is-neither-arithmetic-nor-ge and probably many more. – Gerry Myerson Jan 09 '20 at 18:18
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1Does this answer your question? How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? – Hans Lundmark Jan 09 '20 at 18:35
4 Answers
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Since $$\sum_{n=1}^{\infty}\frac{n}{2^n}=\sum_{n=0}^{\infty}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}+\sum_{n=0}^\infty\frac{1}{2^{n+1}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n}}+1$$
implies that
$$\sum_{n=1}^{\infty}\frac{n}{2^n}=2$$
we can use the same trick to see that
\begin{align}\sum_{n=1}^{\infty}\frac{n^2}{2^n}=\sum_{n=0}^{\infty}\frac{(n+1)^2}{2^{n+1}}&=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n^2}{2^{n}}+\sum_{n=1}^{\infty}\frac{n}{2^{n}}+\sum_{n=0}^\infty\frac{1}{2^{n+1}}\\&= \frac{1}{2}\sum_{n=1}^{\infty}\frac{n^2}{2^{n}}+3 \end{align}
therefore $$\sum_{n=0}^{\infty}\frac{n^2}{2^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{2^{n}}=6$$
Axion004
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Write $c^k$ as the forward difference of $c^k/(c-1)$ with $c=1/2.$ Find the sum from $1$ to $n$ of the given expression by using the summation by parts formula twice. Then take the limit as $n$ goes to infinity.
Allawonder
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P. Lawrence
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The forward difference operator on a function f(x) is defined by Df(x)=f(x+1)-f(x). The summation by parts formula relates the sum of fDg(x) f rom 1 to n to the sum of (D(f))g(x+1) and f(n+1)g(n+1) -f(1)g(1)I. The formula is analogous to integration by parts but there is shifting involved. – P. Lawrence Jan 09 '20 at 17:39
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Start with $\sum_{k=0}^\infty x^k=\frac{1}{1-x}$ for $|x|\lt 1$
$\frac{d}{dx}\frac{1}{1-x}=\sum_{k=0}^\infty kx^{k-1}=\frac{1}{(1-x)^2}$
$\frac{d}{dx}\frac{x}{(1-x)^2}=\sum_{k=0}^\infty k^2x^{k-1}=\frac{1+x}{(1-x)^3}$
Therefore $\sum_{k=0}^\infty k^2x^k=\frac{x(1+x)}{(1-x)^3}$
The question is the value of this last expression when $x=\frac{1}{2}$. Ans $=6$.
herb steinberg
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I guess you've made a mistake.When taking the first derivative, the summation should start from $n=1$. Point is there is no $\frac1x$ term here. Also the summation is $6$. – sbp Jan 09 '20 at 19:45
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My error (corrected} was in the third line derivative. Starting the sum at 0 or 1 doesn't matter when there is a positive power of k as a coefficient, since that term is 0. – herb steinberg Jan 10 '20 at 01:24
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$$\sum_{k=0}^{\infty} x^k=\frac{1}{1-x}, |x|<1.$$ Differentate w.r. $x$, then $$\sum_{k=0}^{\infty} k x^k =\frac{x}{(1-x)^2}.$$ Differentiate again w.r. t. $x$, we get By putting $x=1/2$, we get $$\sum_{k=0}^{\infty} k^2 x^k= \frac{x(1+x)}{(1-x)^3}.$$ $$\sum_{k=0}^{\infty} \frac{k^2}{2^k}=6$$
Z Ahmed
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