let w_n = Σ_(k=2) ^n (1/(k^2 −1)) find lim_(n→∞) w


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Question Number 30524 by abdo imad last updated on 22/Feb/18

$l e t w_{n} = \sum_{k = 2}^{n} \frac{1}{k^{2} - 1} f i n d {l i m}_{n \to \infty} w_{n} .$
Commented by abdo imad last updated on 23/Feb/18

$t h e s e r i e w_{n} i s p o s i t i f h o w t o f i n d a n e g a t i f v a l u e s i r ? . . .$
Commented by abdo imad last updated on 24/Feb/18

$w e h a v e w_{n} = \sum_{k = 2}^{n} \frac{1}{k^{2} - 1} = \frac{1}{2} \sum_{k = 2}^{n} (\frac{1}{k - 1} - \frac{1}{k + 1})$ $= \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k - 1} - \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k + 1}$ $= \frac{1}{2} \sum_{k = 1}^{n - 1} \frac{1}{k} - \frac{1}{2} \sum_{k = 3}^{n + 1} \frac{1}{k} b u t w e h a v e$ $\sum_{k = 1}^{n - 1} \frac{1}{k} = H_{n - 1} a n d \sum_{k = 3}^{n + 1} \frac{1}{k} = H_{n + 1} - \frac{3}{2} \Rightarrow$ $w_{n} = \frac{1}{2} H_{n - 1} - \frac{1}{2} H_{n + 1} + \frac{3}{4} = \frac{1}{2} (H_{n - 1} - H_{n + 1}) + \frac{3}{4}$ $H_{n - 1} = l n (n - 1) + γ + o (\frac{1}{n})$ $H_{n + 1} = l n (n + 1) + γ + o (\frac{1}{n}) \Rightarrow H_{n - 1} - H_{n + 1} = l n (\frac{n - 1}{n + 1}) + o (\frac{1}{n})$ $\Rightarrow {l i m}_{n \to \infty} H_{n - 1} - H_{n + 1} = 0 \Rightarrow {l i m}_{n \to \infty} w_{n} = \frac{3}{4} .$
	Answered by sma3l2996 last updated on 23/Feb/18

	$w_{n} = \underset{k = 2}{\sum^{n}} \frac{1}{k^{2} - 1} = \underset{k = 2}{\sum^{n}} \frac{1}{(k - 1) (k + 1)}$ $w e h a v e \frac{1}{(k + 1) (k - 1)} = \frac{1}{2} (\frac{1}{k - 1} - \frac{1}{k + 1})$ $s o w_{n} = \frac{1}{2} (\underset{k = 2}{\sum^{n}} \frac{1}{k - 1} - \underset{k = 2}{\sum^{n}} \frac{1}{k + 1})$ $i = k - 1; j = k + 1$ $w_{n} = \frac{1}{2} (\underset{i = 1}{\sum^{n - 1}} \frac{1}{i} - \underset{j = 3}{\sum^{n + 1}} \frac{1}{j})$ $w_{n} = \frac{1}{2} (\underset{i = 1}{\sum^{n - 1}} \frac{1}{i} - (\underset{j = 1}{\sum^{n - 1}} \frac{1}{j} + \frac{1}{n} + \frac{1}{n + 1} - 1 - \frac{1}{2}))$ $w_{n} = \frac{1}{2} (\underset{i = 1}{\sum^{n - 1}} \frac{1}{i} - \underset{j = 1}{\sum^{n - 1}} \frac{1}{j} + \frac{2 n + 1}{n (n + 1)} - \frac{3}{2})$ $w_{n} = \frac{1}{2} (\frac{2 n + 1}{n (n + 1)} - \frac{3}{2})$ $Double subscripts: use braces to clarify$
	Commented by sma3l2996 last updated on 24/Feb/18

	$w_{n} = \frac{1}{2} (\underset{i = 1}{\sum^{n - 1}} \frac{1}{i} - (\underset{j = 1}{\sum^{n - 1}} \frac{1}{j} + \frac{2 n + 1}{n (n + 1)} - \frac{3}{2}))$ $= \frac{1}{2} (\underset{i = 1}{\sum^{n - 1}} \frac{1}{i} - \underset{j = 1}{\sum^{n - 1}} - \frac{2 n + 1}{n (n + 1)} + \frac{3}{2})$ $w_{n} = \frac{1}{2} (\frac{3}{2} - \frac{2 n + 1}{n (n + 1)})$ $Double subscripts: use braces to clarify$
	Commented by sma3l2996 last updated on 24/Feb/18

	$I d i d m i s t a k e o n l i n e 7$
	Commented by prof Abdo imad last updated on 24/Feb/18

	$n e v e r m i n d s i r . . .$
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