find-the-value-of-n-2-3n-2-1-n-2-1-3-

Question Number 41236 by maxmathsup by imad last updated on 04/Aug/18

f i n d t h e v a l u e o f \sum_{n = 2}^{\infty} \frac{3 n^{2} + 1}{{(n^{2} - 1)}^{3}}

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Aug/18

{(n + 1)}^{3} - {(n - 1)}^{3} = (n^{3} + 3 n^{2} + 3 n + 1) - (n^{3} - 3 n^{2} + 3 n - 1)

= 6^{} n^{2} + 2

\underset{n = 2}{\sum^{\infty}} \frac{1}{2} . \frac{{(n + 1)}^{3} - {(n - 1)}^{3}}{{(n + 1)}^{3} {(n - 1)}^{3}}

= \frac{1}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{{(n - 1)}^{3}} - \frac{1}{{(n + 1)}^{3}} = \frac{1}{2} \underset{2}{\sum^{\infty}} T_{n}

T_{n} = [\frac{1}{{(n - 1)}^{3}} - \frac{1}{{(n + 1)}^{3}}]

T_{2} = \frac{1}{1^{3}} - \frac{1}{3^{3}}

T_{3} = \frac{1}{2^{3}} - \frac{1}{4^{3}}

T_{4} = \frac{1}{3^{3}} - \frac{1}{5^{3}}

T_{5} = \frac{1}{4^{3}} - \frac{1}{6^{3}}

\dots .

\dots .

T_{n} = \frac{1}{{(n - 1)}^{3}} - \frac{1}{{(n + 1)}^{3}}

s = \frac{1}{1^{3}} + \frac{1}{2^{3}} - \frac{1}{{(n + 1)}^{3}}

\frac{1}{2} \underset{2}{\sum^{\infty}} T_{n} = \frac{1}{2} (1 + \frac{1}{8}) - \lim_{n \to \infty} \frac{1}{{(n + 1)}^{3}} = \frac{9}{16} - 0 = \frac{9}{16}

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Commented by prof Abdo imad last updated on 04/Aug/18

y o u r a n s w e r i s c o r r e c t s i r T a n m a y t h a n k s

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Aug/18

t h a n k y o u s i r \dots

Answered by prof Abdo imad last updated on 04/Aug/18

S = {l i m}_{n \to + \infty} S_{N} w i t h S_{N} = \sum_{n = 2}^{N} \frac{3 n^{2} + 1}{{(n - 1)}^{3} {(n + 1)}^{3}}

b u t w e h a v e {(n + 1)}^{3} - {(n - 1)}^{3}

= 2 ({(n + 1)}^{2} + (n + 1) (n - 1) + {(n - 1)}^{3})

= 2 (n^{2} + 2 n + 1 + n^{2} - 1 + n^{2} - 2 n + 1)

= 2 (3 n^{2} + 1) \Rightarrow S_{N} = \frac{1}{2} \sum_{n = 2}^{N} \frac{{(n + 1)}^{3} - {(n - 1)}^{3}}{{(n - 1)}^{3} {(n + 1)}^{3}}

= \frac{1}{2} {\sum_{n = 2}^{N} \frac{1}{{(n - 1)}^{3}} - \sum_{n = 2}^{N} \frac{1}{{(n + 1)}^{3}}} b u t

\sum_{n = 2}^{N} \frac{1}{{(n - 1)}^{3}} = \sum_{n = 1}^{N - 1} \frac{1}{n^{3}} = ξ_{N - 1} (3)

\sum_{n = 2}^{N} \frac{1}{{(n + 1)}^{3}} = \sum_{n = 3}^{N + 1} \frac{1}{n^{3}} = ξ_{N + 1} (3) - 1 - \frac{1}{8} \Rightarrow

S_{N} = \frac{1}{2} {ξ_{N - 1} (3) - ξ_{N + 1} (3) + \frac{9}{8}} b u t

{l i m}_{N \to + \infty} ξ_{N - 1} (3) - ξ_{N + 1} (3) = ξ (3) - ξ (3) = 0 \Rightarrow

{l i m}_{N \to + \infty} S_{N} = \frac{9}{16} = S