What is the series $\sum_{i=n}^{\infty}\frac{i}{2^i}$ When computing the expected value for a random variable I reached the following series:
$$\sum_{i=n}^{\infty}\frac{i}{2^i}$$
I am confident it is convergent, but have no idea how to compute it.

There are a few standard tricks that one can use.  For example:
Let $\displaystyle f(x) = \sum_{i=n}^\infty i x^{i-1} = \frac{d}{dx} \sum_{i=n}^\infty x^i = \frac{d}{dx} \frac{x^n}{1-x} = \frac{nx^{n-1}(1-x) + x^n }{(1-x)^2}$.
So $\displaystyle \sum_{i=n}^\infty \frac{i}{2^i} = \frac{1}{2} \sum_{i=n}^\infty i \left(\frac{1}{2}\right)^{i-1} = \frac{1}{2} f\left(\frac{1}{2}\right) = \frac{1}{2} \frac{n(1/2)^{n-1} (1/2)+(1/2)^n}{1/4} = \frac{n+1}{2^{n-1}}$.
