Find the value of sum (n/2^n)

Question

I have the series $\sum_{n=0}^\infty \frac{n}{2^n}$. I must show that it converges to 2.

I was given a hint to take the derivative of $\sum_{n=0}^\infty x^n$ and multiply by $x$ , which gives

$\sum_{n=1}^\infty nx^n$ , or $\sum_{n=0}^\infty nx^n$.

Clearly if I take $x=\frac{1}{2}$ , the series is $\sum_{n=0}^\infty \frac{n}{2^n}$. How do I proceed from here?

Start from the identity $\sum_{n \ge 0} x^n = \dfrac1{1-x}$ which holds good for $x \in (0, 1)$, then follow the hint (on both sides of the identity). — Macavity, Apr 06 '16 at 12:46
Perfect, thanks! I was able to solve the problem with this identity, — Dr. John A Zoidberg, Apr 06 '16 at 12:54
Have a look here: http://math.stackexchange.com/questions/337937/why-sum-k-1-infty-frack2k-2 — Martin Sleziak, Apr 06 '16 at 13:30
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, Nov 18 '16 at 06:11
Just a comment, but this is the expected time to get first head of a fair coin. — user135520, Sep 06 '24 at 23:54

score 9 · Accepted Answer · answered Apr 06 '16 at 12:48

9

Notice that if $|x|<1$ then the original series converges with

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

Computing the derivative and plugging in $x=\frac{1}{2}$ should hopefully seem easier now.

answered Apr 06 '16 at 12:48

Mnifldz

1 Answers1