let S_n =Σ_(k=1) ^n (k^2 /((2k−1)(2k+1))) 1) determine S_n interms of n 2) find lim_(n→+∞) S


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Question Number 48493 by maxmathsup by imad last updated on 24/Nov/18

$l e t S_{n} = \sum_{k = 1}^{n} \frac{k^{2}}{(2 k - 1) (2 k + 1)}$ $1) d e t e r m i n e S_{n} i n t e r m s o f n$ $2) f i n d {l i m}_{n \to + \infty} S_{n}$
Commented by maxmathsup by imad last updated on 25/Nov/18

$1) f i r s t l e t d e c o m p o s e F (x) = \frac{x^{2}}{(2 x - 1) (2 x + 1)} \Rightarrow$ $F (x) = \frac{x}{4} (\frac{1}{2 x - 1} + \frac{1}{2 x + 1}) = \frac{1}{8} (\frac{2 x}{2 x - 1} + \frac{2 x}{2 x + 1})$ $= \frac{1}{8} (\frac{2 x - 1 + 1}{2 x - 1} + \frac{2 x + 1 - 1}{2 x + 1}) = \frac{1}{8} (1 + \frac{1}{2 x - 1} + 1 - \frac{1}{2 x + 1})$ $= \frac{1}{4} + \frac{1}{8} (\frac{1}{2 x - 1} - \frac{1}{2 x + 1}) \Rightarrow S_{n} = \frac{n}{4} + \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{2 k - 1} - \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{2 k + 1}$ $b u t \sum_{k = 1}^{n} \frac{1}{2 k + 1} =_{k + 1 = j} \sum_{j = 2}^{n + 1} \frac{1}{2 j - 1} = \sum_{j = 1}^{n} \frac{1}{2 j - 1} - 1 + \frac{1}{2 n + 1} \Rightarrow$ $S_{n} = \frac{n}{4} + \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{2 k - 1} - \frac{1}{8} (\sum_{k = 1}^{n} \frac{1}{2 k - 1} - 1 + \frac{1}{2 n + 1})$ $= \frac{n}{4} + \frac{1}{8} - \frac{1}{8 (2 n + 1)}$ $2) i t s c l e a r t h a t {l i m}_{n \to + \infty} S_{n} = + \infty .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18

	$1) \frac{k^{2}}{(2 k + 1) (2 k - 1)}$ $= \frac{1}{4} [\frac{4 k^{2} - 1 + 1}{(2 k + 1) (2 k - 1)}]$ $= \frac{1}{4} [1 + \frac{1}{2} \times \frac{(2 k + 1) - (2 k - 1)}{(2 k + 1) (2 k - 1}]$ $T_{k} = \frac{1}{4} + \frac{1}{8} [\frac{1}{2 k - 1} - \frac{1}{2 k + 1}]$ $T_{1} = \frac{1}{4} + \frac{1}{8} [\frac{1}{1} - \frac{1}{3}]$ $T_{2} = \frac{1}{4} + \frac{1}{8} [\frac{1}{3} - \frac{1}{5}]$ $T_{3} = \frac{1}{4} + \frac{1}{8} [\frac{1}{5} - \frac{1}{7}]$ $. . .$ $. . .$ $T_{n} = \frac{1}{4} + \frac{1}{8} [\frac{1}{2 n - 1} - \frac{1}{2 n + 1}]$ $a d d t h e m$ $S_{n} . = \frac{n}{4} + \frac{1}{8} [1 - \frac{1}{2 n + 1}]$ $w h e n n \to \infty$ $S_{n} = \infty + \frac{1}{8} [1 - 0] = \infty$ $p l s c h e c k . . .$
	Commented by maxmathsup by imad last updated on 25/Nov/18

	$c o r r e c t a n s w e r t h a n k s .$
	Answered by ajfour last updated on 25/Nov/18

	$S_{n} = \underset{k = 1}{\sum^{n}} \frac{k^{2}}{(2 k - 1) (2 k + 1)}$ $= \frac{1}{4} Σ \frac{k [(2 k + 1) + (2 k - 1)]}{(2 k - 1) (2 k + 1)}$ $4 S_{n} = Σ \frac{k}{2 k - 1} + Σ \frac{k}{2 k + 1}$ $8 S_{n} = Σ (1 + \frac{1}{2 k - 1}) + Σ (1 - \frac{1}{2 k + 1})$ $= 2 n + (\frac{1}{1} + \frac{1}{3} + . . + \frac{1}{2 n - 1})$ $- (\frac{1}{3} + \frac{1}{5} + . . . + \frac{1}{2 n + 1})$ $S_{n} = \frac{n}{4} + \frac{1}{8} (1 - \frac{1}{2 n + 1}) .$
	Commented by maxmathsup by imad last updated on 25/Nov/18

	$c o r r e c t a n s w e r t h a n k s$
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