# Compute the limit $\sum_{n=1}^{\infty} \frac{n}{2^n}$ [closed]

Compute the limit $$\sum_{n=1}^{\infty} \frac{n}{2^n}$$

• I found it! Your turn! Mar 12, 2013 at 10:20
• See this post and its linked pages for ideas. Mar 12, 2013 at 10:37
• Me too, Asaf! Hint to OP: Do not use imperative when asking a question here. It's a violation of site etiquette. May 12, 2015 at 1:31

Hint: Consider the power series with those coefficients, for $x=1$. Remember that we can integrate term-by-term within the convergence radius.

• Ha-Ha! I was first :) Mar 12, 2013 at 10:22
• Yes, but I'm a set theorist. And I don't have to teach calculus this semester! So... jokes on someone else. Mar 12, 2013 at 10:22

By ratio test, $\displaystyle \lim_{x\to \infty}\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^{n}}}=\lim_{x\to \infty}\frac{n+1}{2n}=\frac 1 2<1$.

Therefore $\displaystyle\sum_{n=1}^\infty \frac n {2^n}$ convergent.

$\displaystyle S=\sum_{n=1}^\infty \frac n {2^n}=\frac 1 2 + \frac 2 {2^2} + \frac 3 {2^3}+...$ (i)

$2*$(i): $\displaystyle 2S=1 + \frac 2 {2} + \frac 3 {2^2}+...$ (ii)

(ii) $-$ (i): $\displaystyle S=1+\frac 1 2 + \frac 1 {2^2} +...$

Can you take it from here? (Answer: $S=2$)

p.s. This method requires just elementary skills. :)

• Maybe it should be added that this works if we know that the series converges (i.e., $S\ne+\infty$). Mar 12, 2013 at 13:30
• @ Martin Sleziak: You are right, otherwise, $\infty-\infty$ is indeterminate.
– JSCB
Mar 12, 2013 at 13:32
• If we subtract $S$ from $2S$, how can we conclude that the difference is equal to $S$ ? May 11, 2015 at 18:57

Hints:
(1) $f(x)=\frac1{1-x}=\sum_{n=0}^\infty x^n$ converges uniformly for $|x|\leq r$ for any $r<1$ and hence is differentiatable.
(2) $f'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac1x\sum_{n=1}^\infty nx^{n}$

• Hint: If you use 1. the software parses it as an ordered list (you may have to line break one more) Mar 12, 2013 at 10:22