calculate S_p =Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)(n+2)...(n+p))) with p fromN


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 41409 by maxmathsup by imad last updated on 06/Aug/18

$c a l c u l a t e S_{p} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n (n + 1) (n + 2) . . . (n + p)} w i t h p f r o m N$
	Answered by sma3l2996 last updated on 07/Aug/18

	$\frac{1}{n (n + 1) (n + 2) . . . (n + p)} = \frac{a_{0}}{n} + \frac{a_{1}}{n + 1} + . . . + \frac{a_{p}}{n + p}$ $w i t h a_{i} = \underset{n \to - i}{l i m} \frac{n + i}{n (n + 1) . . . (n + p)}$ $\frac{1}{n (n + 1) (n + 2) . . . (n + p)} = \underset{i = 0}{\sum^{p}} \frac{a_{i}}{n + i}$ $s o S_{p} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n (n + 1) (n + 2) . . . (n + p)} = \underset{n = 1}{\sum^{\infty}} \underset{i = 0}{\sum^{p}} \frac{{(- 1)}^{n} a_{i}}{n + i}$ $= \underset{i = 0}{\sum^{p}} a_{i} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + i}$ $l e t m = n + i$ $S_{n} = \underset{i = 0}{\sum^{p}} a_{i} \underset{m = i + 1}{\sum^{\infty}} \frac{{(- 1)}^{m - i}}{m}$ $= \underset{i = 0}{\sum^{\infty}} a_{i} {(- 1)}^{- i} (\underset{m = i + 1}{\sum^{\infty}} \frac{{(- 1)}^{m}}{m} + \underset{m = 1}{\sum^{i}} \frac{{(- 1)}^{m}}{m} - \underset{m = 1}{\sum^{i}} \frac{{(- 1)}^{m}}{m})$ $= \underset{i = 0}{\sum^{p}} {(- 1)}^{i} a_{i} (\underset{m = 1}{\sum^{\infty}} \frac{{(- 1)}^{m}}{m} - \underset{m = 1}{\sum^{i}} \frac{{(- 1)}^{m}}{m})$ $n o t e : \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} = - l n (2)$ $T h e r e f o r e :$ $S_{p} = \underset{i = 0}{\sum^{p}} {(- 1)}^{i} a_{i} (- l n (2) - \underset{m = 1}{\sum^{i}} \frac{{(- 1)}^{m}}{m})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum