# $\sum_{n=0}^{\infty} n \left( 0.5 \right)^n$ [closed]

How to find the sum of the following series:

$$\sum_{n=0}^{\infty} n \left( 0.5 \right)^n$$

## closed as off-topic by Davide Giraudo, Claude Leibovici, Tom Oldfield, drhab, Najib IdrissiJul 13 '14 at 15:10

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• This post might help. – David Mitra Jul 13 '14 at 13:08
• @DavidMitra Might help? There is a solution there! – m0nhawk Jul 13 '14 at 13:10
• Why the downvote – Vishwa Iyer Jul 13 '14 at 13:16
• @DavidMitra, thank you for the link. – Yal dc Jul 13 '14 at 13:20
• This is a polylogarithm. – Lucian Jul 13 '14 at 18:36

## 3 Answers

\begin{align}(0.5)&+(0.5)^2&+(0.5)^3+\cdots\\ &+(0.5)^2&+(0.5)^3+\cdots\\&&+(0.5)^3+\cdots\end{align}

• Then I'll end up summing $2 + (2- 0.5) + (2-0.5-0.5^2) ...$. I don't see how that became easier. – Yal dc Jul 13 '14 at 13:09
• Look again. The first row is 1, the second is 1/2, the third is 1/4... – Empy2 Jul 13 '14 at 13:11
• Actually, I looked in the answers from the link provided by Mr.Mitra and now I understand what you have in mind. – Yal dc Jul 13 '14 at 13:12

Let $$S = \sum_{n = 0}^{\infty} n(0.5)^n$$ $$S = 0 + 0.5 + 2*(0.5)^2 + 3*(0.5)^3 ....$$ $$0.5S = 0 + (0.5)^2 + 2*(0.5)^3 + 3*(0.5)^4 ....$$ $$S - 0.5S = 0 + 0.5 + (0.5)^2 + (0.5)^3 ...$$ As you can recall, this is the sum of an infinite geometric series with $r = \frac{1}{2}$. So $$0.5S = \frac{0.5}{1-0.5}$$ $$S = 2$$

We know that (from generating functions or geometric series): $$\frac 1{1-x} = \sum_{k=0}^\infty x^k$$ differentiating both sides: $$\frac1{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k$$ Now subtracting and setting $x=\frac 12$: $$\sum_{k=0}^\infty k\left(\frac 12\right)^k = \frac 1{\left(1-\frac 12\right)^2}-\frac 1{1-\frac 12}=4-2=2$$