The question is Σ_(n=1) ^∞ (((2n−1)/(n(n+1)(n+2)))=...


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Question Number 103633 by Lordose last updated on 16/Jul/20

$The question is$ $\underset{n = 1}{\sum^{\infty}} (\frac{2 n - 1}{n (n + 1) (n + 2}) = . . .$
Commented by bobhans last updated on 16/Jul/20

$w e l l w e l l$
	Answered by Dwaipayan Shikari last updated on 16/Jul/20
	$S=Σ^∞ (((n+n+2−3)/(n(n+1)(n+2))))=Σ^∞ (1/((n+1)(n+2)))+Σ^∞ (1/(n(n+1)))−3S_1 S=Σ^∞ (1/(n+1))−(1/(n+2))+Σ^∞ (1/n)−(1/(n+1))−3S_1 =(3/2)−3S_1 S_1 =(1/2)Σ^∞ ((2+n−n)/(n(n+1)(n+2)))=(1/2)Σ^∞ (1/n)−(1/(n+1))−(1/2)Σ^∞ (1/(n+1))−(1/(n+2)) S_1 =(1/2)−(1/4)=(1/4) { Σ^∞ (1/n)−(1/(n+1))=lim_(n→∞) 1−(1/(n+1))=1 S=(3/2)−(3/4)=(3/4) {Σ^∞ (1/(n+1))−(1/(n+2))=lim_(n→∞) (1/2)−(1/(n+2))=(1/2)$
	$S = \sum^{\infty} (\frac{n + n + 2 - 3}{n (n + 1) (n + 2)}) = \sum^{\infty} \frac{1}{(n + 1) (n + 2)} + \sum^{\infty} \frac{1}{n (n + 1)} - 3 S_{1}$ $S = \sum^{\infty} \frac{1}{n + 1} - \frac{1}{n + 2} + \sum^{\infty} \frac{1}{n} - \frac{1}{n + 1} - 3 S_{1} = \frac{3}{2} - 3 S_{1}$ $S_{1} = \frac{1}{2} \sum^{\infty} \frac{2 + n - n}{n (n + 1) (n + 2)} = \frac{1}{2} \sum^{\infty} \frac{1}{n} - \frac{1}{n + 1} - \frac{1}{2} \sum^{\infty} \frac{1}{n + 1} - \frac{1}{n + 2}$ $S_{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} {\sum^{\infty} \frac{1}{n} - \frac{1}{n + 1} = \underset{n \to \infty}{\lim 1} - \frac{1}{n + 1} = 1$ $S = \frac{3}{2} - \frac{3}{4} = \frac{3}{4} {\sum^{\infty} \frac{1}{n + 1} - \frac{1}{n + 2} = \lim_{n \to \infty} \frac{1}{2} - \frac{1}{n + 2} = \frac{1}{2}$
	Commented by mathmax by abdo last updated on 16/Jul/20

	$not correct$
	Commented by Dwaipayan Shikari last updated on 16/Jul/20

	$kindly can you show my mistake ?$
	Answered by Worm_Tail last updated on 16/Jul/20

	$\underset{n = 1}{\sum^{\infty}} (\frac{2 n - 1}{n (n + 1) (n + 2}) = A$ $p f d$ $A = Σ (\frac{3}{n + 1} - \frac{5}{2 (n + 2)} - \frac{1}{2 n})$ $A = 3 Σ (\frac{1}{n + 1}) - \frac{5}{2} Σ (\frac{1}{n + 2}) - \frac{1}{2} Σ (\frac{1}{n})$ $A = 3 (\frac{1}{2} + \sum_{n = 2} (\frac{1}{n + 1})) - \frac{5}{2} \sum_{n = 1} (\frac{1}{n + 2}) - \frac{1}{2} (1 + \frac{1}{2} + \sum_{n = 3} (\frac{1}{n}))$ $\sum_{n = 2} (\frac{1}{n + 1}) = \sum_{n = 1} (\frac{1}{n + 2}) = \sum_{n = 3} (\frac{1}{n}) = s$ $A = 3 (\frac{1}{2} + s) - \frac{5}{2} s - \frac{1}{2} (1 + \frac{1}{2} + s)$ $A = \frac{3}{2} + 3 s - \frac{5}{2} s - \frac{1}{2} - \frac{1}{4} - \frac{1}{2} s$ $A = 3 s - \frac{5}{2} s - \frac{1}{2} s + \frac{3}{4}$ $A = \frac{3}{4}$
	Commented by mathmax by abdo last updated on 16/Jul/20

	$correct answer thanks$
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