calculus - 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?

Answer

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

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