[sweetjazz]

My favorite summation trick using the geometric series is the following argument:

sum of x^n = (1 - x)^(-1)
differentiate to get
sum of n x^(n-1) = (1 - x)^(-2)
multiply by x to get
sum of n x^n = x * (1 - x)^(-2)

So, e.g.
sum of n / 2^n = 2.