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How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$

How can you compute the limit of $\sum \limits_{n=1}^{\infty} n(2/3)^n$

Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?

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marked as duplicate by Eric Naslund, GEdgar, t.b., Jonas Meyer, Asaf Karagila Dec 18 '11 at 21:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Hint: For $|x|< 1$, try expressing $\frac{1}{1-x}$ as a series in $x$. Then differentiate both sides with respect to $x$, and set $x = \frac{2}{3}$ in the new inequality. – Dilip Sarwate Dec 12 '11 at 1:17
Related question:… (In particular, this answer gives a more general result for finite sum – Martin Sleziak Dec 13 '11 at 12:10
@user9352 Honestly I can't figure out what is the meaning of "Evidently it is equal to $6$ by Wolfram Alpha"... It seems like "software always got all the right answers" and the latter is a sound nonsense. – Pacciu Dec 18 '11 at 12:54

6 Answers 6

up vote 15 down vote accepted

$$ \begin{align*} \sum_{n=1}^\infty n(2/3)^n &= \sum_{m=1}^\infty \sum_{n=m}^\infty (2/3)^n \\ &= \sum_{m=1}^\infty \frac{(2/3)^m}{1-2/3} \\ &= \frac{2/3}{(1-2/3)^2} = 6. \end{align*} $$

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+1. It may be worth to add, for clarity: $\sum_{n=1}^\infty n x^n = \sum_{n=1}^\infty \sum_{m=1}^n x^n = \sum_{m=1}^\infty \sum_{n=m}^\infty x^n$. – Sasha Dec 12 '11 at 1:36
I think the indexing was off - it looked like you had a non-convergent inner sum. I took the liberty of editing, but please check. – alex.jordan Dec 12 '11 at 8:21

You can do this with power series. If you let $f(x) := \sum \limits_{n=1}^{\infty} nx^n$ and restrict the domain of $f$ to the interval $|x|<1$ then $$\begin{align} f(x) &= x \sum_{n=1}^{\infty} nx^{n-1} \\&= x \sum_{n=1}^{\infty} \frac{d}{dx} x^n \\&= x\frac{d}{dx}(\sum_{n=1}^{\infty} x^n) \\&= x \frac{d}{dx} \bigg( \frac{1}{1-x}\bigg) \\&= \frac{x}{(1-x)^2} \end{align} $$ and substituting $x=2/3$ gives $f(x)=6$.

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Toss a coin that has probability $1/3$ of landing "heads" until we get a head. Let $X$ be the number of tosses required. We find the mean of $X$ in two different ways. Let our sum be $S$.

Note that $P(X=1)=1/3$, $P(X=2)=(2/3)(1/3)$, $P(X=3)=(2/3)^2(1/3)$, and so on. It follows that $$E(X)=1\cdot\left(\frac{1}{3}\right)+ 2\cdot\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)+ 3\cdot\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)+ 4\cdot\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)+\cdots.$$ Thus $$E(X)=\frac{1}{2}\left[1\cdot\left(\frac{2}{3}\right)+ 2\cdot\left(\frac{2}{3}\right)^2+ 3\cdot\left(\frac{2}{3}\right)^3+ 4\cdot\left(\frac{2}{3}\right)^4+\cdots\right]=\frac{S}{2}.$$

If the first toss is a head, then $X=1$. If that the first toss is a tail, we have used up $1$ toss, and the game begins again. By the Law of Total Expectation, $$E(X)=1\cdot\frac{1}{3} +(1+E(X))\cdot\frac{2}{3}.$$ Solve for $E(X)$. We get $E(X)=3$, and therefore $S=6$.

Comment: There is a very nice book on bijective arguments called Proofs that Really Count. Maybe one should start collecting Mean Proofs.

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very interesting, sometimes I wonder, how does this world looks in the eyes of people like yourself? It must be very different from my world! Also, thanks for naming the book. – NoChance Dec 18 '11 at 6:21
@Emmad Kareem: We are both doing mathematics. I have been doing it a lot longer. – André Nicolas Dec 18 '11 at 6:24
very humble-Thanks. – NoChance Dec 18 '11 at 6:34
This made my day! Thanks! – Hidde Dec 18 '11 at 10:14

Almost the same as Mike's answer:

Let $$ \def\ts{\textstyle} S_n=\ts{2\over3}+2( {2\over 3})^2 +3 ({2\over3})^3+\cdots+n ({2\over3})^n.$$

Then $$\eqalign{ \ts{2\over3}S_n &=\ts \bigl[\,({2\over3})^2+2 ({2\over 3})^3+3 ({2\over3})^4 +\cdots+(n-1) ({2\over3})^{n }\,\bigr]+n ({2\over3})^{n+1}\cr &=\ts S_n- [{2\over3}+ ({2\over3})^2 +({2\over3})^3 + \cdots + ({2\over3})^{n } ] + n({2\over3})^{n +1} \cr &=\ts S_n-{2/3 -(2/3)^{n+1}\over 1/3}+ n({2\over3})^{n +1}. }$$


$$S_n={ {2/3 -(2/3)^{n+1}\over 1/3}- n({2\over3})^{n +1}\over 1/3}.$$

Taking the limit as $n$ tends to infinity gives $$ S_n={2-0\over 1/3}=6. $$

This method was observed (for the general differentiated geometric series $\sum n r^n$) by Roger B. Nelsen, who has another lovely proof here.

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Also see this Math Stack Exchange answer which does what you are talking about for general $r$: – Eric Naslund Dec 18 '11 at 13:04

I've read about doing it like this:

$y=\sum \limits_{n=1}^\infty n(\frac23)^n$

$\frac23y=\sum \limits_{n=1}^\infty n(\frac23)^{n+1}=\sum \limits_{n=2}^\infty (n-1)(\frac23)^n=\sum \limits_{n=1}^\infty (n-1)(\frac23)^n$, since $n=1$ yields $0$.

$y-\frac23y=\sum \limits_{n=1}^\infty [n-(n-1)](\frac23)^n$

$\frac13y=\sum \limits_{n=1}^\infty (\frac23)^n$

This sum you may be more familiar with. If not, you can solve it similarly to obtain



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Just a curioursity.

Using Fourier series one comes to: $$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx = 6\ \frac{13 \cos x-12}{(13-12\cos x)^2}$$ hence: $$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n = \sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx \Bigg|_{x=0} =6\ \frac{13 \cos x-12}{(13-12\cos x)^2}\Bigg|_{x=0} =6\; .$$

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