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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) 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

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}} + \dots + \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 \dots

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}]

\dots

\dots . n o w l o o k w h e n w e a d d T_{2} + T_{3} + T_{4} + \dots

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 \dots

Commented by Tawa1 last updated on 06/Jan/19

God bless you sir

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