# Evaluate $\sum_{n=1}^\infty \frac{n}{2^n}$.

Evaluate $$\sum_{n=1}^\infty \frac{n}{2^n}$$

My Work:

$$\sum_{n=1}^\infty \frac{n}{2^n} = \sum_{n=1}^\infty n \left(\frac{1}{2}\right)^n$$

If we denote $f(x) = \sum_{n=1}^\infty nx^n$ then we wish to evaluate $f(1/2)$.

Now, $$\sum_{n=1}^\infty nx^n = x \sum_{n=1}^\infty nx^{n-1} = x\sum_{n=1}^\infty (x^n)' = x\left(\sum_{n=1}^\infty x^n\right)' = x \left(\frac{-x}{1-x}\right)' = \frac{-x}{(1-x)^2}$$

Applying $x=1/2$ we get the wrong result of $-2$.

Where is my mistake?

The mistake was in the fourth step -- the $-x$ in $\left(\frac{-x}{1 - x}\right)^{'}$ should be $x$. So you should have

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

• But $$\sum_{n=1}^\infty x^n = \frac{-x}{1-x}$$ – AlonAlon Mar 3 '15 at 16:40
• @AlonAlon no, it's $\sum_{n = 1}^\infty x^n = \frac{x}{1 - x}$. – kobe Mar 3 '15 at 16:41
• It's because $$(1 - x)\sum_{n = 1}^N x^n = \sum_{n = 1}^N (x^n - x^{n+1}) = x - x^{N+1} \to x$$ which implies $$(1 - x)\sum_{n = 1}^\infty x^n = x$$ or $$\sum_{n = 1}^\infty x^n = \frac{x}{1 - x}$$ – kobe Mar 3 '15 at 16:44

Yet another solution (maybe less tricky).

You can express this as a double summation and exchange the order (this is directly related to the integration by parts): \begin{align} \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}{2^n} 1_{k \leq n}\\ & = \sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{2^n} 1_{k \leq n}\\ & = \sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{2^n}\\ & = \sum_{k=1}^\infty \frac{1}{2^k}\frac{1}{1-\frac{1}{2}}\\ & = 2\sum_{k=1}^\infty \frac{1}{2^k}\\ & = 2. \end{align}

• What $1_{k\le n}$ is? – AlonAlon Mar 3 '15 at 16:46
• This is just a notation $1_{k \leq n} = \begin{cases}1 & \text{if }k\leq n\\0 & \text{otherwise}\end{cases}$. – Siméon Mar 3 '15 at 16:47