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

Question Number 38520 by math khazana by abdo last updated on 26/Jun/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 - 1)}^{3} {(n + 1)}^{3}}

Commented by abdo mathsup 649 cc last updated on 28/Jun/18

l e t S_{n} = \sum_{k = 2}^{n} \frac{3 k^{2} + 1}{{(k - 1)}^{3}) {(k + 1)}^{3}}

w e h a v e {(k + 1)}^{3} - {(k - 1)}^{3}

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

= 2 {k^{2} + 2 k + 1 + k^{2} - 1 + k^{2} - 2 k + 1}

= 2 {3 k^{2} + 1} \Rightarrow

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

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

= \frac{1}{2} \sum_{k = 1}^{n - 1} \frac{1}{k^{3}} - \frac{1}{2} \sum_{k = 3}^{n + 1} \frac{1}{k^{3}}

= \frac{1}{2} ξ_{n - 1} (3) - \frac{1}{2} {ξ_{n + 1} (3) - 1 - \frac{1}{8}}

= \frac{1}{2} {ξ_{n - 1} (3) - ξ_{n + 1} (3)} + \frac{9}{16} b u t

{l i m}_{n \to \infty} ξ_{n - 1} (3) = ξ (3)

{l i m}_{n \to \infty} ξ_{n + 1} (3) = ξ (3) \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{9}{16} .