Question

(a) Starting with the geometric series [infinity] xn n = 0 , find the sum of the series [infinity] nxn − 1 n = 1 , |x| < 1.

Answer 1

Let f(x) be the sum of the geometric series,

`f(x)=\displaystyle\frac1{1-x} = \sum_(n=0)^\infty x^n`

for |x| < 1. Then taking the derivative gives the desired sum,

`f'(x)=\displaystyle\boxed{\frac1{(1-x)^2}} = \sum_(n=0)^\infty nx^(n-1) = \sum_(n=1)^\infty nx^(n-1)`