3

I am stuck on the following problem:

I have to determine the sum of the series $$\sum_{n=1}^{\infty}\frac{n+1}{2^n}$$

My Attempt: $$\sum_{n=0}^{\infty}\frac{n+1}{2^n}=\sum_{n=0}^{\infty}\frac{1}{2^n}+\sum_{n=0}^{\infty}\frac{n}{2^n}=\frac{1}{1-\frac12}+\sum_{n=0}^{\infty}\frac{n}{2^n}=2+\sum_{n=0}^{\infty}\frac{n}{2^n}$$.
So,I am stuck on determining the value of $\,\,\sum_{n=0}^{\infty}\frac{n}{2^n}$.

Can someone point me in the right direction? Thanks in advance for your time.

learner
  • 6,726
  • Wow, even Wolfram|Alpha can't show steps for this one. – Ahaan S. Rungta Nov 06 '13 at 15:28
  • See this. – David Mitra Nov 06 '13 at 15:37
  • While you're at it, you could try to find the sum of $\sum (n+1)x^n$ for $|x|<1$. Hint: it is a derivative. – Julien Nov 06 '13 at 15:38
  • 1
    $ \begin{array}{l}if:x = \frac{1}{2} \Rightarrow \sum\limits_{n = 0}^{ + \infty } {\left( {\frac{1}{2}} \right)^n } + \sum\limits_{n = 0}^{ + \infty } {n\left( {\frac{1}{2}} \right)^n } = \frac{1}{{1 - \frac{1}{2}}} + \frac{{\frac{1}{2}}}{{\left( {1 - \frac{1}{2}} \right)^2 }} = 2 + 2 = 4 \ \end{array} $ – sabaga Jun 11 '14 at 19:23

3 Answers3

4

$$\sum_{n=1}^\infty \frac{n}{2^n} = \sum_{n=1}^\infty \sum_{k=1}^n \frac{1}{2^n}$$ $$ = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1_{(k \leq n)}}{2^n}$$ $$= \sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{2^n}$$ $$ = \sum_{k=1}^\infty \frac{\frac{1}{2^k}}{1-.5}$$ $$ = \sum_{k=1}^\infty \frac{1}{2^{k-1}}$$ $$= 2$$

Gautam Shenoy
  • 10,318
1

Hint: Denote $S=\sum_{n=0}\frac{n}{2^n}$. What would it be $2S-S$?

Shuchang
  • 9,800
0

Take the geometric series $f(x)=\sum_{n=1}^\infty x^{n+1}$. Look at its derivative and plug in $x=1/2$. Of course you need to justify that everything works, but that's standard.

primoz
  • 206