Let f(x) be the sum of the geometric series,
[tex]f(x)=\displaystyle\frac1{1-x} = \sum_{n=0}^\infty x^n[/tex]
for |x| < 1. Then taking the derivative gives the desired sum,
[tex]f'(x)=\displaystyle\boxed{\dfrac1{(1-x)^2}} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}[/tex]
