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This question already has an answer here:

Apparently,

$$ \sum_{n = 0}^\infty \frac{n}{2^n} $$

converges to 2. I'm trying to figure out why. I've tried viewing it as a geometric series, but it's not quite a geometric series since the numerator increases by 1 every term.

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marked as duplicate by Jyrki Lahtonen, Ayman Hourieh, Start wearing purple, Dan Rust, Micah Jul 11 '13 at 20:04

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.

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    $\begingroup$ If you're not too worried about rigor, here's a nice trick: take the geometric series formula $(1 - \alpha)^{-1} = \sum_0^{\infty} \alpha^k$ and differentiate in $\alpha$ once on either side. Now evaluate at $\alpha = 1/2$. $\endgroup$ – A Blumenthal Jul 11 '13 at 18:39
  • $\begingroup$ @ABlumenthal , is your reasoning only possible if $a_n$ converges uniformly? I don't remember the exact condition for taking the derivative on both sides of the equation. $\endgroup$ – sigmatau Jul 11 '13 at 18:42
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    $\begingroup$ It's enough for the derivative series to converge uniformly, and the original series to converge somewhere. $\endgroup$ – Chris Eagle Jul 11 '13 at 18:43
  • $\begingroup$ see math.stackexchange.com/questions/92224 $\endgroup$ – sdcvvc Jul 11 '13 at 18:48

10 Answers 10

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Besides the differentiation trick mentioned by others, here's another trick:

$$S = \sum_{n=0}^{\infty} \frac{n}{2^n} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{n}{2^{n-1}} = \frac{1}{2} \left(\sum_{n=0}^{\infty} \frac{n - 1}{2^{n-1}} + \sum_{n=0}^{\infty} \frac{1}{2^{n-1}}\right) = \frac{1}{2} \left(S + \frac{-1}{2^{-1}} + 4\right) = \frac{1}{2}(S + 2).$$

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$$\begin{array}{} \sum_{n\ge 0}\frac{n}{2^n}&=&\frac1{2^1}&+&\frac2{2^2}&+&\frac3{2^3}&+&\frac4{2^4}&+&\ldots&=\\ \hline &&\frac1{2^1}&+&\frac1{2^2}&+&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 1}\left(\frac12\right)^n=1\\ &&&&\frac1{2^2}&+&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 2}\left(\frac12\right)^n=\frac12\\ &&&&&&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 3}\left(\frac12\right)^n=\frac14\\ &&&&&&&&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 4}\left(\frac12\right)^n=\frac18\\ &&&&&&&&&&\ddots&\vdots&\qquad\vdots\\ &&&&&&&&&&&&\color{blue}{\sum_{n\ge 0}\frac1{2^n}=2} \end{array}$$

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    $\begingroup$ This way of thinking about it can also be made clear by drawing a suitably shaped figure. If you split the area into vertical strips of unit width, you see that the total area is the sum on Brian's rightmost column. OTOH if you split it into horizontal strips of heights $2^{-n}$ and width $n$ you see that the total area is the original sum. $\endgroup$ – Jyrki Lahtonen Jul 11 '13 at 19:14
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Hint: $$\begin{align} \frac12+\frac14+\frac18+\frac1{16}+\dots&=\color{red}{1}\\ \frac14+\frac18+\frac1{16}+\dots&=\color{red}{\frac12}\\ \frac18+\frac1{16}+\dots&=\color{red}{\frac14}\\ \frac1{16}+\dots&=\color{red}{\frac18}\\ \hline \frac12+\frac24+\frac38+\frac4{16}+\dots&=2 \end{align}$$

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    $\begingroup$ Thanks for your support. :+) $\endgroup$ – mrs Jul 13 '13 at 19:04
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This is the derivative of a geometric series: Let $$f(x)=\sum_{n=0}^\infty x^n.$$ Then (by taking derivative summandwise) $$f'(x) = \sum_{n=1}^\infty nx^{n-1}.$$ Since $f(x)=\frac1{1-x}$ if $|x|<1$, we have $f'(x)=\frac1{(1-x)^2}$. Your sum is just $\frac12f'(\frac12)$.

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Consider the power series $$f(x)=\sum_{n=0}^\infty\frac{x^n}{2^n}=\sum_{n=0}\left(\frac x2\right)^n=\sum_{n=0}^\infty t^n=\frac{1}{1-t}=\frac{1}{1-\frac x2}=\frac{2}{2-x}$$ Then we have that $f(1)=2$. We also have that $$f'(x)=\sum_{n=0}^\infty\frac{nx^{n-1}}{2^n}$$

But we also have that $$\left(\frac2{2-x}\right)'=\frac2{(2-x)^2}$$

Therefore $f'(1)=2$ as wanted.

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    $\begingroup$ I'll be glad to hear what's wrong in this answer. $\endgroup$ – Asaf Karagila Jul 11 '13 at 20:08
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The method that says $$ \sum_n nx^n = x \sum_n \frac{d}{dx} x^n,\text{ etc, etc.,} $$ has already been mentioned.

Here's another way: $$ \begin{array}{rrrrrrrrrrrrrrrr} 1/2 \\[8pt] {}+ 1/4 & {}+ 1/4 \\[8pt] {}+ 1/8 & {}+1/8 & {}+1/8 \\[8pt] {}+\cdots{} \end{array} $$

Now find the sum of the entries in the first column. Then the second column. Then the third. And so on.

After that, find the sum of those sums.

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Toss a fair coin until you get a head. Let $X$ be the number of tosses. Your sum is $E(X)$. Write $a$ for $E(X)$.

With probability $\frac{1}{2}$ you get a head on the first toss. Given this happened, $E(X)=1$.

With probability $\frac{1}{2}$, you get a tail on the first toss. Given this happened, $E(X)=1+a$.

Thus $$a=\frac{1}{2}\cdot 1+\frac{1}{2}\cdot a.$$ Solve for $a$.

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Think about, in general,

$$S = 1+2 r+3 r^2 + 4 r^3 + \cdots$$

$$r S = r + 2 r^2 + 3 r^3 + \cdots$$

$$S - r S = (1-r) S = 1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r}$$

Therefore

$$S = \frac{1}{(1-r)^2}$$

This, however, is not really the series you listed; that series has an extra factor of $r$; so the sum is actually $r/(1-r)^2$. Plug in $r=1/2$ and the sum is $2$.

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You know that $\sum_{k=0}^\infty \frac{1}{2^k}=2$ from the geometric series formula. Write \begin{align*} 1&= \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots \\ \frac{1}{2}&=0 +\frac{1}{2^2}+\frac{1}{2^3}+\cdots \\ \frac{1}{4}&=0+0+\frac{1}{2^3}+\cdots \\ \vdots&= \qquad\ddots \end{align*}

Summing the total of each column on the right and using the known fact that the left hand side is $2$, you get your desired answer without any other tricks

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  • $\begingroup$ Why the downvote? $\endgroup$ – TMM Jul 11 '13 at 20:21
  • $\begingroup$ @TMM Perhaps because my array is not perfectly aligned? :O $\endgroup$ – Jean-Sébastien Jul 11 '13 at 20:25
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Lets call your series $\Sigma$.

Lets seperate the $\Sigma$ into a sub series we'll call ${1 \over 2} \Sigma$.

${1 \over 2}\Sigma$ is a sub series of $\Sigma$, right?

Lets subtract ${1 \over 2} \Sigma$ from $\Sigma$. Will call the result as $S_{1 \over 2}$.

Well, $S_{1 \over 2} = {1 \over 2}\Sigma - \Sigma:$

$$\Sigma = {1 \over 2} + {2 \over 4} + {3 \over 8} + {4 \over 16} + \dots$$

as $${1 \over 2} \Sigma = {1 \over 4} + {2 \over 8} + {3 \over 16} + {4 \over 32} + \dots $$

We've devided and subtracted half series from the original series, $\Sigma$,and got $S_{1 \over 2}$,

Well, $\Sigma$ is $2 {S_{1 \over 2}}$. (Why?)

We can see that ${S_{1 \over 2}}$ = $\Sigma$ - ${1 \over 2}\Sigma$ $=>$

$${1 \over 2} + {2 \over 4} + {3 \over 8} + {4 \over 16} + \dots$$ $$ - $$ $${1 \over 4} + {2 \over 8} + {3 \over 16} + {4 \over 32} + \dots $$

$$ = $$

$$ {1 \over 2} - {1 \over 4} + {2 \over 4} - {2 \over 8} + {3 \over 8} - {3 \over 16} + {4 \over 16} - {4 \over 18} \dots$$

which is:

$$ {1 \over 2} + {1 \over 4} + {1 \over 8} + {1 \over 16} + \dots$$

Can you notice what happened? We got a Geometric series.

Hence, we can easly get the convertange limit of $S_{1 \over 2}$: ${a_1} \over {1 - q}$ , in our case:

$$S_{1\over2} = {{1 \over 2} \over {1 - {1 \over 2}}} = {{1 \over 2} \over {1 \over 2}} = 1$$

Wait! We've devided by 2. Well, it's time to turn back :) multiply $S_{1 \over 2}$ by the inverse of ${1 \over 2}$, which is 2.

Well, $\Sigma = 2S_{1\over2} = 2*1 = 2$.

Done!

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