# 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 + … ?

Sep 25, 2014

Alternating Harmonic Series

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} / n = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$

Since

$\ln \left(1 - x\right) = - {\sum}_{n = 1}^{\infty} \frac{{x}^{n}}{n}$,

by setting $x = - 1$,

$\ln 2 = - {\sum}_{n = 1}^{\infty} \frac{{\left(- 1\right)}^{n}}{n} = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} / n$