calculate-S-p-n-1-1-n-n-n-1-n-2-n-p-with-p-fromN-

Question Number 41409 by maxmathsup by imad last updated on 06/Aug/18

c a l c u l a t e S_{p} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n (n + 1) (n + 2) \dots (n + p)} w i t h p f r o m N

Answered by sma3l2996 last updated on 07/Aug/18

\frac{1}{n (n + 1) (n + 2) \dots (n + p)} = \frac{a_{0}}{n} + \frac{a_{1}}{n + 1} + \dots + \frac{a_{p}}{n + p}

w i t h a_{i} = \underset{n \to - i}{l i m} \frac{n + i}{n (n + 1) \dots (n + p)}

\frac{1}{n (n + 1) (n + 2) \dots (n + p)} = \underset{i = 0}{\sum^{p}} \frac{a_{i}}{n + i}

s o S_{p} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n (n + 1) (n + 2) \dots (n + p)} = \underset{n = 1}{\sum^{\infty}} \underset{i = 0}{\sum^{p}} \frac{{(- 1)}^{n} a_{i}}{n + i}

= \underset{i = 0}{\sum^{p}} a_{i} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + i}

l e t m = n + i

S_{n} = \underset{i = 0}{\sum^{p}} a_{i} \underset{m = i + 1}{\sum^{\infty}} \frac{{(- 1)}^{m - i}}{m}

= \underset{i = 0}{\sum^{\infty}} a_{i} {(- 1)}^{- i} (\underset{m = i + 1}{\sum^{\infty}} \frac{{(- 1)}^{m}}{m} + \underset{m = 1}{\sum^{i}} \frac{{(- 1)}^{m}}{m} - \underset{m = 1}{\sum^{i}} \frac{{(- 1)}^{m}}{m})

= \underset{i = 0}{\sum^{p}} {(- 1)}^{i} a_{i} (\underset{m = 1}{\sum^{\infty}} \frac{{(- 1)}^{m}}{m} - \underset{m = 1}{\sum^{i}} \frac{{(- 1)}^{m}}{m})

n o t e : \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} = - l n (2)

T h e r e f o r e :

S_{p} = \underset{i = 0}{\sum^{p}} {(- 1)}^{i} a_{i} (- l n (2) - \underset{m = 1}{\sum^{i}} \frac{{(- 1)}^{m}}{m})