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How do you prove $$\sum_{n=1}^{\infty} \frac {n}{2^n} = 2\ ?$$

My attempt: I have been trying to find geometric series that converge to 2 which can bind the given series on either side. But I am unable to find these. Is there a general technique to find the sum? This is a high school interview question and must be easy enough to solve in a few minutes.

Please give any hints for the first step towards a solution.


marked as duplicate by Hans Lundmark, vadim123, mathlove, user223391, N. F. Taussig Jun 18 '15 at 19:47

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    $\begingroup$ Can you use $\sum_n x^n = {1 \over 1-x}$ for $|x|<1$, differentiate and set $x={1 \over 2}$? $\endgroup$ – copper.hat Jun 18 '15 at 18:04
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    $\begingroup$ related : this $\endgroup$ – mathlove Jun 18 '15 at 18:07

Hint: Denote your sum by $S$, rewrite $S=2S-S$ and try to rewrite the sums.


One approach I find intuitive and avoids derivatives:

$$\sum_{n=1}^{\infty} \frac {n}{2^n} = \sum_{n=1}^{\infty} \frac {1}{2^n} + \sum_{n=2}^{\infty} \frac {1}{2^n} + \sum_{n=3}^{\infty} \frac {1}{2^n} + \cdots $$

Now analyze this.

  • $\begingroup$ Excellent! That's great, this should have all the up votes! $\endgroup$ – Zach466920 Jun 18 '15 at 18:09
  • $\begingroup$ @SimonS Just curious ... how is this different from Hans Engler's outlined approach? $\endgroup$ – Mark Viola Jun 18 '15 at 21:35
  • $\begingroup$ @Dr.MV They're the same, although I believe I posted first. $\endgroup$ – Simon S Jun 18 '15 at 22:23

Here is a simple way to see this. It ignores technical aspects of rearranging infinite series.

Write down the series as follows: $$ \quad \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \dots \\ = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\ \qquad + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\ \qquad \qquad \quad + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\ \qquad \qquad + \frac{1}{2^4} + \dots \\ \vdots $$ Each row is a geometric series. The values of the rows are $1, \frac{1}{2}, \, \frac{1}{2^2}, \frac{1}{2^3}$, which is again a geometric series. Add this series and you'll get the answer $\boxed{2}$.


A general way to handle similar problems is to note that

$$\sum_{n=1}^{\infty}nx^n=x\frac{d}{dx}\sum_{n=1}^{\infty}x^n \tag1$$

The sum on the right-hand side of $(1)$ is the series for

$$\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}\tag 2$$

Taking a derivative of the right-hand side of $(2)$ reveals that


whereupon evaluation at $x=1/2$ provide the expected result


  • $\begingroup$ Why did this just receive a down vote? $\endgroup$ – Mark Viola Jul 22 '16 at 12:42

Notice that $\dfrac{1}{1-x} = \sum_{i=0}^\infty x^i$.

Now differentiate: $\dfrac{1}{(1-x)^2} = \sum_{i=1}^\infty ix^{i-1}$.

You asked for a hint, but there isn't much left to do.

  • $\begingroup$ Op asked for a hint...also he's in high school so derivatives might work... $\endgroup$ – Zach466920 Jun 18 '15 at 18:05
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    $\begingroup$ Lay off the downvoting $\endgroup$ – copper.hat Jun 18 '15 at 18:05
  • $\begingroup$ @copper.hat I guess a hint is just a solution now a days ;) also I'll vote whichever way I please, thank you very much, but I do also listen to the public. $\endgroup$ – Zach466920 Jun 18 '15 at 18:08

Here's my hint.

This is very similar to the geometric sum $$S=\sum_{n=0}^{\infty} a^n$$ simply try expanding the terms out and analyze the term by term expansion of $$S-S \cdot a$$ and look at the derivation for the geometric sum.


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