Sum of a given series

**Jame** · March 9th 2013, 11:16 AM

Hello

I was wondering if there was any way to calculate the sum of the series

$\sum_{k=1}^\infty \frac{k}{2^k}$

I know it is convergent by the ratio test, but it there anyway to finding out what sum it converges to?

This series came up when I was calculating the expected value of a random variable.

(Wolfram Alpha says it is 2, which makes sense in context of the problem)

**Lockdown** · March 9th 2013, 12:51 PM

You can prove that the sum converges to 2 by considering the partial sums.
It turns out that (I encourage you to prove this):
$\sum_{k=1}^n \frac{k}{2^k} = \frac{2^{n+1}-n-2}{2^n}$ .

Hence, taking the limit as n tends to infinity, you can see that the sum converges to 2.

**Soroban** · March 9th 2013, 12:59 PM

Hello, Jame!

I was wondering if there was any way to calculate the sum of the series

. . $\sum_{k=1}^\infty \frac{k}{2^k}$

$\begin{array}{ccccc}\text{We have: }& S & =& \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \cdots \\ \\[-4mm] \text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}S &=& \quad\;\; \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \cdots \\ \\[-4mm] \text{Subtract: }& \frac{1}{2}S &=& \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \cdots \end{array}$

The right side is a geometric series
. . with first term $a = \tfrac{1}{2}$ and common ratio $r = \tfrac{1}{2}.$
Its sum is: . $\frac{\frac{1}{2}}{1-\frac{1}{2}} \:=\:1$

Therefore: . $\tfrac{1}{2}S \:=\:1 \quad\Rightarrow\quad S \:=\:2$

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