-2

A user just posted a similar question (What is the summation of $\sum_{k=0}^{\infty}\frac{k}{2^k}$?) with $n$ instead of $n^2$, and this question got some original answers, so I would like to propose the question:

Find the sum $$\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}.$$

I am really interested to see what original answers are given to this question. There are number of different approaches, some I havent seen or though of before. Thus although the question may be unoriginal, some of the solutions are not. Thus it is inappropriate to close this question.

Rene Schipperus
  • 39,526

5 Answers5

2

By stars and bars we have $$ \frac{1}{(1-x)^{k+1}}=\sum_{n\geq 0}\binom{n+k}{k} x^n $$ for any $x\in(-1,1)$, in particular $$ \sum_{n\geq 0}\binom{n+k}{k}\frac{1}{2^n} = 2^{k+1} $$ for any $k\in\mathbb{N}$. Since $n^2 = 2\binom{n+2}{2}-3\binom{n+1}{1}+\binom{n+0}{0}$ we have $$ \sum_{n\geq 0}\frac{n^2}{2^n} = 2\cdot 2^3 - 3\cdot 2^2 + 2^1 = \color{red}{6}. $$

Jack D'Aurizio
  • 353,855
  • Sir, I have two questions, first, I know how to prove $\displaystyle \sum_{n\ge 0}\binom{n+k}kx^n=\frac1{(1-x)^{k+1}}$ using binomial theorem, but how stars and bars? Secondly, how did you know that $\displaystyle n^2 = 2\binom{n+2}{2}-3\binom{n+1}{1}+\binom{n+0}{0}$, was it just a manipulation to convert it to the form $\displaystyle \binom{n+k}k$ or is there a specific way in which we can write? For example, how do we write $n^3$ as the sum of such binomial coefficients? – V.G Feb 25 '21 at 13:43
1

We know that $$\sum_{n\geq 0} r^n = \frac{1}{1-r}$$ Now, differentiate both sides: $$\sum_{n\geq 0} \left(r^{n}\right) = \left(\frac{1}{1-r}\right)'$$ $$\sum_{n\geq 0} nr^{n-1} = \frac{1}{(1-r)^2}$$ Differentiating again: $$\sum_{n\geq 0} (nr^{n-1})' = \left(\frac{1}{(1-r)^2}\right)'$$ $$\sum_{n\geq 0} n(n-1)r^{n-2} = \frac{2}{(1-r)^3}$$ $$\sum_{n\geq 0} n^2r^{n-2}-\sum_{n\geq 0} nr^{n-2} = \frac{2}{(1-r)^3}$$ $$\sum_{n\geq 0} n^2r^{n-2} =\sum_{n\geq 0} nr^{n-2}+ \frac{2}{(1-r)^3}$$ Now multiple both sides by $r^2$: $$\sum_{n\geq 0} n^2r^{n} =\sum_{n\geq 0} nr^{n}+ \frac{2r^2}{(1-r)^3}$$ $$\sum_{n\geq 0} n^2r^{n} =\frac{r}{(1-r)^2}+ \frac{2r^2}{(1-r)^3}$$

Botond
  • 11,938
  • 5
    Did you copy-paste your other answer and just insert the second diffreentiation? ;) – Hagen von Eitzen Apr 14 '18 at 13:00
  • @HagenvonEitzen Yes, I think it's better to copy it than just to refer to it :) – Botond Apr 14 '18 at 13:02
  • Botond Not so sure: if you can essentially copy and paste an answer to an earlier question, then we can consider the question you paste your answer to, a duplicate of the one you already answered. There is such a thing known as "duplicate answers"... when you knowingly add one line to an answer you already posted, you are admitting it's essentially a dupe of the first, and hence you should vote to close as a dupe, and not try to copy and paste your answers to yet another question of the same variety. – amWhy Apr 14 '18 at 21:29
  • @amWhy I don't agree with you now. It's a bit different question, not the same. Yes, I was using the same approach. That's why I copied the beginning of my answer. I could also say that "Check this answer, but instead of differentiating it one, do it twice and multiply it by $r^2$ now.". But that would not be a quality answer. It's quite similar to answering an $\epsilon-\delta$ limit question, and at the next time, copying the same "for all..." definition to the beginning. – Botond Apr 14 '18 at 21:44
  • I see this as simply double dipping; the OP here is fully familiar with the answers to the question you copied from. I think ethically, you should have pursued your second alternative: Link, and suggest a modification. That you answered instead of helping to trace the many duplicates of the question, reflects poorly on you as well. – amWhy Apr 14 '18 at 21:50
  • And this answer which this answer resembles strikingly. – amWhy Apr 14 '18 at 22:08
  • @amWhy I did what I thought as the right way. Yes, I could do the way you mentioned. I get your point, but I don't regret it. You can downvote and/or flag my answer (or even delete it, if you can, I can't see your permissions) if you'd like to, I will not be angry at all. But as you can see, I did copy only 3 row of equations and extended it with 6 more. So it's not just a full copy. And I think you can use your knowledge more than once in this site. Or are you referring to your previous answers every time, like "Here, I'm using something you can find in this answer of mine..."? – Botond Apr 14 '18 at 22:18
  • @amWhy I don't know where did I learn this technique. Who knows, it could be from one of your links. Or from a teacher of mine. But I think neither I nor you will find it out. – Botond Apr 14 '18 at 22:21
  • No, I have no plans to dv or delete your answer (no one can singlehandedly delete a post save for a mod), nor flag it. Just asking you to consider searching (even if just skimming down the column of "related posts" we see to the right of the question/answer fields) if the question has already been asked and answered. Not saying dedicating much time to searching; but over time you will develop a good feel of what has likely (or surely) been asked before. I'm not scolding you, or blaming you for anything. It's just an issue to be aware of. – amWhy Apr 14 '18 at 22:22
  • @amWhy Thanks for telling me your honest opinion, I appreciate it. I will pay more attention to this next time. – Botond Apr 14 '18 at 22:33
1

\begin{align*} \sum\limits_{n=1}^{N}\frac{(n+1)^2}{2^n}-\sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=\sum\limits_{n=1}^{N}\frac{2n+1}{2^n}\\ 2\sum\limits_{n=1}^{N}\frac{(n+1)^2}{2^{n+1}}-\sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{N}\frac{n}{2^n}+\sum\limits_{n=1}^{N}\frac{1}{2^n}\\ 2\sum\limits_{n=2}^{N}\frac{n^2}{2^n}-\sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{N}\frac{n}{2^n}+\sum\limits_{n=1}^{N}\frac{1}{2^n}\\ 2\left(\sum\limits_{n=1}^{N}\frac{n^2}{2^n}-\frac{1}{2}+\frac{(N+1)^2}{2^{N+1}}\right)-\sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{N}\frac{n}{2^n}+\sum\limits_{n=1}^{N}\frac{1}{2^n}\\ \sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{N}\frac{n}{2^n}+\sum\limits_{n=1}^{N}\frac{1}{2^n}+1-\frac{(N+1)^2}{2^N}\\ \sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{\infty}\frac{n}{2^n}+\sum\limits_{n=1}^{\infty}\frac{1}{2^n}+1\\ &=2(2)+(1)+1\\ &=6 \end{align*}

CY Aries
  • 23,393
  • Kinda tricky to read all this algebra, since you're doing unexplained operations on both sides. Do you think you could maybe replace some of the sums with variables instead of writing them out in full (since you use $\sum n^2/2^n$ at least 5 times) or perhaps explaining your algebra between each step? – Jam Apr 14 '18 at 13:11
1

Using a probabilistic interpretation as the second moment $E(X^2)$ of a random varaible $X$ with a geometric distribution $\text{Geo}(p=1/2$) (beginning at $1$ !), it suffices to know (https://en.wikipedia.org/wiki/Geometric_distribution) its mean and its variance, resp. :

$$E(X)=\tfrac{1}{p} \ \ \text{and} \ \ E(X^2)-E(X)^2=\tfrac{1-p}{p^2}$$

to be able to conclude that:

$$E(X^2)=(\tfrac{1}{p})^2+\tfrac{1-p}{p^2}=\tfrac{2-p}{p^2} \ \ \text{with} \ \ p=\tfrac12,$$ finally giving the sum : $6.$

Jean Marie
  • 81,803
1

Following my telescoping sum strategy from the similar question mentioned in your question -

$$\sum_{k=0}^m \left( \frac{(k+1)^2}{2^{k+1}} - \frac{k^2}{2^k} \right) = \frac{(m+1)^2}{2^{m+1}}$$

$$ \Longrightarrow \sum_{k=0}^m \frac{-k^2+2k+1}{2^{k+1}} = \frac{(m+1)^2}{2^{k+1}}$$

$$ \Longrightarrow \frac{1}{2} \sum_{k=0}^m \frac{k^2}{2^k} = \sum_{k=0}^m \frac{k}{2^k} + \frac{1}{2} \sum_{k=0}^m \frac{1}{2^k} - \frac{(m+1)^2}{2^{m+1}}$$

And using the partial summation for the first term of the RHS (found in my answer to the similar question), we have

$$\sum_{k=0}^m \frac{k^2}{2^k} = 2 \left( 2- \frac{m+2}{2^m} \right) + \left( 2-\frac{1}{2^m} \right) - \frac{(m+1)^2}{2^m}$$

and taking the limit as $m$ approaches infinity yields

$$\sum_{k=0}^{\infty} \frac{k^2}{2^k} = 6.$$

user328442
  • 2,705