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