Sum of $\sum_{n=1}^{\infty}\frac{n}{2^n}$.

Question

I was trying to find the sum of $\sum_{n=1}^{\infty}\frac{n}{2^n}$.

I tried like this $S = \sum_{n=1}^{\infty}\frac{n}{2^n} = \sum_{n=1}^{\infty}(\frac{n+1}{2^n} - \frac{1}{2^n}) = 2\sum_{n=2}^{\infty}\frac{n}{2^n} - 1$

So,$S = 2(S - \frac{1}{2}) - 1$

Implying $S =2$.

EDIT:

Also from many posts in the comment below I found a method of looking at the sum $\sum x^n$ and differentiating and plugging for $x$.

Is there any other method of looking at the problem and calculating the sum apart from the above method,may be it would be interesting to see different approaches to the same problem which may be used to visualize other problems based on summation of series?

https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn — user236182, Sep 24 '17 at 11:04
One might note some ambiguity with your sigma notation. I'd prefer to write$$\sum_{n=1}^\infty\left(\frac{n+1}{2^n}-\frac1{2^n}\right)=\sum_{n=1}^\infty\left(\frac{n+1}{2^n}\right)-1$$ — Simply Beautiful Art, Sep 24 '17 at 11:53
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$?, Why does $\sum_{n = 0}^\infty \frac{n}{2^n}$ converge to 2?, What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to?. If you search in approach0 you can find more questions about the same sum. — Martin Sleziak, Sep 24 '17 at 12:12

score 2 · Accepted Answer · answered Sep 24 '17 at 11:24

2

Multiply with $(2-1)$: $$ (2-1)\sum_{n=1}^\infty n2^{-n}=\sum_{n=1}^\infty(n2^{1-n}-n2^{-n})=\sum_{n=0}^\infty (n+1)2^{-n}-\sum_{n=1}^\infty n2^{-n}=\sum_{n=0}^\infty2 ^{-n}$$

answered Sep 24 '17 at 11:24

Hagen von Eitzen

374,180

score 1 · Answer 2 · answered Sep 24 '17 at 11:02

1

Summing $x^n$ gives $\frac{1}{1-x}-1$, so summing $nx^{n-1}$ gives $\frac{1}{(1-x)^2}$. Set $x=1/2$.

answered Sep 24 '17 at 11:02

J.G.

115,835

MrYouMath · Answer 3 · 2017-09-24T11:24:35.170

Hint: Use the geometric series

$$\frac{1}{1-u}=\sum_{n=0}^{\infty}u^n.$$

This series is uniformly convergent for $|u|\leq 1$, hence we can differentiate it with respect to $u$, in which $|u|\leq 1$.

$$-\frac{1}{(1-u)^2}=\sum_{n=1}^{\infty}n^{} u^{n-1} \implies -\frac{u}{(1-u)^2}=\sum_{n=1}^{\infty}n^{} u^{n} .$$

Relate this sum to your sum, in which $u=1/2$. Note, that your series is starting at $n=2$ and this series is starting at $n=1$.

Sum of $\sum_{n=1}^{\infty}\frac{n}{2^n}$.

3 Answers3