# why $\sum_{h=0}^{\infty}{\dfrac{h}{2^h}} = 2$ [duplicate]

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There is a summation in analysis of an algorithm which is the following:

$$\sum_{h=0}^{\infty}{\dfrac{h}{2^h}} = \dfrac{1/2}{(1-1/2)^2} = 2$$

But I don't can't solve this. I would be appreciated if anyone can solve this.

thanks.

## marked as duplicate by lab bhattacharjee, Hans Lundmark, Ian, Masacroso, RohanJan 2 '18 at 14:58

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## 3 Answers

Not sure what you need to solve (there doesn't seem to be an unknown in your expression), but assuming you want to derive the result:

$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $|x|<1$. Differentiate both sides to get $\sum_{n=0}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$. Multiply both sides by $x$ to finally get:

$\sum_{n=0}^{\infty}nx^n=\frac{x}{(1-x)^2}$

use that $$\sum_{i=0}^n\frac{i}{2^i}=-2\, \left( 1/2 \right) ^{n+1} \left( n+1 \right) -2\, \left( 1/2 \right) ^{n+1}+2$$ and compute the Limit for $n$ tends to infinity

Just for fun, here is a way to find the partial sums via telescoping:

First, it is clear that $$\sum_{k=0}^n ( \frac{k+1}{2^{k+1}}-\frac{k}{2^k} ) = \frac{n+1}{2^{n+1}}.$$

So, rewriting the LHS yields (just some simple algebra)

$$\sum_{k=0}^n (\frac{-1}{2} \cdot \frac{k}{2^k} + \frac{1}{2^{k+1}} ) = \frac{n+1}{2^{n+1}}.$$

From here, we can separate the sum and rearrange the equation to get

$$\frac{-1}{2} \cdot \sum_{k=0}^n \frac{k}{2^k} = \frac{n+1}{2^{n+1}} - \frac{1}{2} \cdot \sum_{k=0}^n \frac{1}{2^k}.$$

If we know the closed form of the geometric sum (to clean up the RHS) ahead of time then we have

$$\sum_{k=0}^n \frac{k}{2^k} = -\frac{n+1}{2^n} - \frac{1}{2^n} + 2.$$

Taking the limit as $n$ approaches infinity yields

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

Note: This technique isn’t the usual way to go about this problem but it can be used to evaluate the sum when higher positive integer powers are in the numerator. For example, I came up with this idea in a competition a couple years to show (2-step process)

$$\sum_{k=0}^{\infty} \frac{k^2}{5^k} = \frac{15}{32}.$$