calculate-S-p-n-0-1-n-n-p-

Question Number 72912 by mathmax by abdo last updated on 04/Nov/19

c a l c u l a t e S_{p} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + p}

Answered by mind is power last updated on 04/Nov/19

S_{p} = \sum_{n ⩾ p} \frac{{(- 1)}^{k - p}}{k}

p ⩾ 2

= {(- 1)}^{p} . \sum_{n ⩾ p} \frac{{(- 1)}^{k}}{k} = {(- 1)}^{p} . (\sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n} - \underset{n = 1}{\sum^{p - 1}} \frac{{(- 1)}^{n}}{n})

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