Question

How do you use a power series to find the exact value of the sum of the series `1-1/2+1/3-1/4 + …` ?

Answer 1

Alternating Harmonic Series

`sum_{n=1}^infty(-1)^{n-1}/n=1-1/2+1/3-1/4+cdots=ln2`

Since

`ln(1-x)=-sum_{n=1}^infty{x^n}/n`,

by setting `x=-1`,

`ln2=-sum_{n=1}^infty{(-1)^n}/n=sum_{n=1}^infty(-1)^{n-1}/n`