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

Question Number 36820 by maxmathsup by imad last updated on 06/Jun/18

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

Commented by math khazana by abdo last updated on 24/Jun/18

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

l e t d e c o m p o s e F (x) = \frac{1}{{(x - 1)}^{2} {(x + 1)}^{2}}

F (x) = \frac{a}{x - 1} + \frac{b}{{(x - 1)}^{2}} + \frac{c}{x + 1} + \frac{d}{{(x + 1)}^{2}}

b = {l i m}_{x \to 1} {(x - 1)}^{2} F (x) = \frac{1}{4}

d = {l i m}_{x \to - 1} {(x + 1)}^{2} F (x) = \frac{1}{4} \Rightarrow

F (x) = \frac{a}{x - 1} + \frac{1}{4 {(x - 1)}^{2}} + \frac{c}{x + 1} + \frac{1}{4 {(x + 1)}^{2}}

{l i m}_{x \to + \infty} x F (x) = 0 = a + c \Rightarrow c = - a \Rightarrow

F (x) = \frac{a}{x - 1} - \frac{a}{x + 1} + \frac{1}{4 {(x - 1)}^{2}} + \frac{1}{4 {(x + 1)}^{2}}

F (0) = 1 = - 2 a + \frac{1}{2} \Rightarrow - 2 a = \frac{1}{2} \Rightarrow a = - \frac{1}{4} \Rightarrow

F (x) = \frac{1}{4 (x + 1)} - \frac{1}{4 (x - 1)} + \frac{1}{4 {(x - 1)}^{2}} + \frac{1}{4 {(x + 1)}^{2}}

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

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

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

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

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

\sum_{k = 2}^{n} \frac{1}{{(k + 1)}^{2}} = \sum_{k = 3}^{n + 1} \frac{1}{k^{2}} = ξ_{n + 1} (2) - \frac{5}{4} \Rightarrow

4 S_{n} = H_{n + 1} - \frac{3}{2} - H_{n - 1} + ξ_{n - 1} (2) + ξ_{n + 1} (2) - \frac{5}{4}

4 S_{n} = γ + l n (n + 1) + o (\frac{1}{n}) - γ - l n (n - 1) - o (\frac{1}{n})

+ ξ_{n - 1} (2) + ξ_{n + 1} (2) - \frac{11}{4} \Rightarrow

4 S_{n} \to \frac{2 π^{2}}{6} - \frac{11}{4} = \frac{π^{2}}{3} - \frac{11}{4} \Rightarrow

{l i m}_{n \to + \infty} S_{n} = \frac{π^{2}}{12} - \frac{11}{16} = S .

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18

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

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

= \frac{1}{4} \times \frac{(n + 1) - (n - 1)}{(n + 1) {(n - 1)}^{2}} - \frac{1}{4} \times \frac{(n + 1) - (n - 1)}{{(n + 1)}^{2} (n - 1)}

= \frac{1}{4} \times \frac{1}{{(n - 1)}^{2}} - \frac{1}{4} \times \frac{1}{(n + 1) (n - 1)} -

\frac{1}{4} \times \frac{1}{(n + 1) (n - 1)} + \frac{1}{4} \times \frac{1}{{(n + 1)}^{2}}

= \frac{1}{4} \times \frac{1}{{(n - 1)}^{2}} - \frac{1}{2} \times \frac{1}{(n + 1) (n - 1)} + \frac{1}{4} \times \frac{1}{{(n + 1)}^{2}}

s o

S_{n} = S_{1} - S_{2} + S_{3} (n > 1)

S_{1} = \frac{1}{4} (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots . .) = \frac{1}{4} \times \frac{\prod^{2}}{6}

S_{2} = \frac{1}{4} {\frac{(n + 1) - (n - 1)}{(n + 1) (n - 1)}}

= \frac{1}{4} {\frac{1}{n - 1} - \frac{1}{n + 1}}

= {\frac{1}{4} (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots) - \frac{1}{4} (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \dots)}

= \frac{1}{4} \times \frac{3}{2} = \frac{3}{8}

S_{3} = \frac{1}{4} \times {\frac{1}{{(n + 1)}^{2}}}

= \frac{1}{4} \times (\frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}} + \dots .)

= \frac{1}{4} \times (\frac{\prod^{2}}{6} - \frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{1}{4} (\frac{\prod^{2}}{6} - \frac{5}{4})

S_{n} = S_{1} - S_{2} + S_{3}

= \frac{1}{4} \times \frac{\prod^{2}}{6} - \frac{1}{4} \times \frac{3}{2} + \frac{1}{4} (\frac{\prod^{2}}{6} - \frac{5}{4})

= \frac{1}{4} \times {\frac{\prod^{2}}{6} - \frac{3}{2} + \frac{\prod^{2}}{6} - \frac{5}{4}}

= \frac{1}{4} \times {\frac{2 \prod^{2}}{6} - \frac{11}{4}}

Commented by math khazana by abdo last updated on 24/Jun/18

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