Determine if the series converges or diverges. (i) Σ_(n = 2) ^∞ (1/(n^2 − 1)) (ii) Σ


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Question Number 52349 by Tawa1 last updated on 06/Jan/19

$Determine if the series converges or diverges .$ $(i) \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2} - 1}$ $(ii) \underset{n = 1}{\sum^{\infty}} \frac{1}{3^{n - 1}}$
Commented by Abdo msup. last updated on 07/Jan/19
$1) let S_n =Σ_(k=2) ^n (1/(k^2 −1)) ⇒S_n =(1/2)Σ_(k=2) ^n ((1/(k−1))−(1/(k+1))) =(1/2)Σ_(k=2) ^n (1/(k−1)) −(1/2)Σ_(k=2) ^n (1/(k+1)) but Σ_(k=2) ^n (1/(k−1)) =Σ_(k=1) ^(n−1) (1/k) Σ_(k=2) ^n (1/(k+1)) =Σ_(k=3) ^(n+1) (1/k) =Σ_(k=1) ^(n−1) (1/k) −(3/2) +(1/n) +(1/(n+1)) ⇒S_n =(1/2){Σ_(k=1) ^(n−1) (1/k) −Σ_(k=1) ^(n−1) (1/k) +(3/2)−(1/n)−(1/(n+1))} =(3/4) −(1/(2n)) −(1/(3n+2)) ⇒lim_(n→+∞) S_n =(3/4) .$
$1) l e t S_{n} = \sum_{k = 2}^{n} \frac{1}{k^{2} - 1} \Rightarrow S_{n} = \frac{1}{2} \sum_{k = 2}^{n} (\frac{1}{k - 1} - \frac{1}{k + 1})$ $= \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k - 1} - \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k + 1} b u t$ $\sum_{k = 2}^{n} \frac{1}{k - 1} = \sum_{k = 1}^{n - 1} \frac{1}{k}$ $\sum_{k = 2}^{n} \frac{1}{k + 1} = \sum_{k = 3}^{n + 1} \frac{1}{k} = \sum_{k = 1}^{n - 1} \frac{1}{k} - \frac{3}{2} + \frac{1}{n} + \frac{1}{n + 1}$ $\Rightarrow S_{n} = \frac{1}{2} {\sum_{k = 1}^{n - 1} \frac{1}{k} - \sum_{k = 1}^{n - 1} \frac{1}{k} + \frac{3}{2} - \frac{1}{n} - \frac{1}{n + 1}}$ $= \frac{3}{4} - \frac{1}{2 n} - \frac{1}{3 n + 2} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{3}{4} .$
Commented by Abdo msup. last updated on 07/Jan/19
$ii)let S_n = Σ_(k=1) ^n (1/3^(k−1) ) =Σ_(k=0) ^(n−1) ((1/3))^k =((1−((1/3))^n )/(1−(1/3))) =(3/2){1−(1/3^n )} ⇒lim_(n→+∞) S_n =(3/2)$
$i i) l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{3^{k - 1}} = \sum_{k = 0}^{n - 1} {(\frac{1}{3})}^{k}$ $= \frac{1 - {(\frac{1}{3})}^{n}}{1 - \frac{1}{3}} = \frac{3}{2} {1 - \frac{1}{3^{n}}} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{3}{2}$
Commented by Tawa1 last updated on 07/Jan/19

$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

	$i i) S_{n} = \frac{1}{3^{0}} + \frac{1}{3} + \frac{1}{3^{2}} + . . . + \frac{1}{3^{n - 1}}$ $S_{n} = \frac{1 (1 - \frac{1}{3^{n}})}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{3^{n}}}{\frac{2}{3}}$ $S = \lim_{n \to \infty} \frac{1 - \frac{1}{3^{n}}}{\frac{2}{3}} = \frac{1 - 0}{\frac{2}{3}} = \frac{3}{2} s o c o n v e r g e . . .$
	Commented by Tawa1 last updated on 06/Jan/19

	$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

	$1) T_{n} = \frac{1}{2} [\frac{(n + 1) - (n - 1)}{(n + 1) (n - 1)}]$ $T_{n} = \frac{1}{2} [\frac{1}{n - 1} - \frac{1}{n + 1}]$ $T_{2} = \frac{1}{2} [\frac{1}{1} - \frac{1}{3}]$ $T_{3} = \frac{1}{2} [\frac{1}{2} - \frac{1}{4}]$ $T_{4} = \frac{1}{2} [\frac{1}{3} - \frac{1}{5}]$ $T_{5} = \frac{1}{2} [\frac{1}{4} - \frac{1}{6}]$ $. . .$ $. . . . n o w l o o k w h e n w e a d d T_{2} + T_{3} + T_{4} + . . .$ $a l l n u m e r c a n c e l l e d e x c e p t \frac{1}{1} a n d \frac{1}{2}$ $s o a n s w d r i s = \frac{1}{2} [1 + \frac{1}{2}] = \frac{3}{4} s o c o n v e r g e . . .$
	Commented by Tawa1 last updated on 06/Jan/19

	$God bless you sir$
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