Σ_(n=1) ^∞ (1/((n+1)(n+2)(n+3)))=


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Question Number 80036 by jagoll last updated on 30/Jan/20

$\underset{n = 1}{\sum^{\infty}} \frac{1}{(n + 1) (n + 2) (n + 3)} =$
Commented by john santu last updated on 30/Jan/20
$lim_(p→∞) Σ_(n=1) ^p (((n+3)−(n+1))/(2(n+1)(n+2)(n+3))) = (1/2)lim_(p→∞) Σ_(n=1 ) ^p (1/((n+1)(n+2)))−(1/((n+2)(n+3))) =(1/2){(1/(2.3))−(1/(3.4))+(1/(3.4))−(1/(4.5))+(1/(4.5))−(1/(5.6))+...} the last term equat to zero , all in between terms are cancelled ∴ Σ_(n=1) ^∞ (1/((n+1)(n+2)(n+3))) = (1/(12))$
$\lim_{p \to \infty} \underset{n = 1}{\sum^{p}} \frac{(n + 3) - (n + 1)}{2 (n + 1) (n + 2) (n + 3)}$ $= \frac{1}{2} \lim_{p \to \infty} \underset{n = 1}{\sum^{p}} \frac{1}{(n + 1) (n + 2)} - \frac{1}{(n + 2) (n + 3)}$ $= \frac{1}{2} {\frac{1}{2.3} - \frac{1}{3.4} + \frac{1}{3.4} - \frac{1}{4.5} + \frac{1}{4.5} - \frac{1}{5.6} + . . .}$ $the last term equat to zero,$ $all in between terms are cancelled$ $∴ \underset{n = 1}{\sum^{\infty}} \frac{1}{(n + 1) (n + 2) (n + 3)} = \frac{1}{12}$
Commented by mr W last updated on 30/Jan/20

$t h e q u e s t i o n w a s o r i g i n a l l y$ $\underset{n = 1}{\sum^{\infty}} \frac{n}{(n + 1) (n + 2) (n + 3)} =$ $w h i c h h a s t h e r e s u l t \frac{1}{4} .$
Commented by jagoll last updated on 31/Jan/20

$w a w . . . i h a v e t w o s o l u t i o n$ $(1) \underset{n = 1}{\sum^{\infty}} \frac{n}{(n + 1) (n + 2) (n + 3)} = \frac{1}{4}$ $(2) \underset{n = 1}{\sum^{\infty}} \frac{1}{(n + 1) (n + 2) (n + 3)} = \frac{1}{12}$ $t h a n k s y o u$
	Answered by Kamel Kamel last updated on 30/Jan/20

	$\frac{n}{(n + 1) (n + 2) (n + 3)} = \frac{A}{n + 1} + \frac{B}{n + 2} + \frac{C}{n + 3} . . . (E)$ $(E) \times (n + 1), n = - 1 \Rightarrow A = - \frac{1}{2}$ $(E) \times (n + 2), n = - 2 \Rightarrow B = 2$ $(E) \times (n + 3), n = - 3 \Rightarrow C = - \frac{3}{2}$ $∴ \frac{n}{(n + 1) (n + 2) (n + 3)} = \frac{1}{2} (\frac{1}{n + 2} - \frac{1}{n + 1}) - \frac{3}{2} (\frac{1}{n + 3} - \frac{1}{n + 2})$ $∴ \underset{n = 1}{\sum^{+ \infty}} \frac{n}{(n + 1) (n + 2) (n + 3)} = \frac{1}{2} (\underset{n = 1}{\sum^{+ \infty}} (\frac{1}{n + 2} - \frac{1}{n + 1}) - 3 \underset{n = 1}{\sum^{+ \infty}} (\frac{1}{n + 3} - \frac{1}{n + 2}))$ $= \frac{1}{2} ((\frac{1}{3} - \frac{1}{2} + \frac{1}{4} - \frac{1}{3} + . . .) - 3 (\frac{1}{4} - \frac{1}{3} + \frac{1}{5} - \frac{1}{4} + . . .))$ $= \frac{1}{2} (- \frac{1}{2} + 1) = \frac{1}{4}$
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