n-1-1-n-n-

Question Number 145951 by ArielVyny last updated on 09/Jul/21

\sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n} = ? ?

Answered by Olaf_Thorendsen last updated on 09/Jul/21

\frac{1}{1 + x} = 1 - x + x^{2} - x^{3} + \dots = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{n}

\ln ∣ 1 + x ∣ = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1} = - \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{n}}{n}

if x = 1

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} = - \ln 2