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Question Number 33713 by abdo imad last updated on 22/Apr/18

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

Commented by prof Abdo imad last updated on 26/Apr/18

l e t p u t S_{n} = \sum_{k = 2}^{n} \frac{n^{2} - n + 1}{{(n - 1)}^{2} {(n + 1)}^{2}} l e t ? d r c o m p o s e

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

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

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

c = {l i m}_{x \to - 1} {(x + 1)}^{2} F (x) = \frac{3}{4} \Rightarrow

F (x) = \frac{a}{x - 1} + \frac{1}{4 {(x - 1)}^{2}} + \frac{b}{x + 1} + \frac{3}{4 {(x + 1)}^{2}}

F (0) = 1 = - a + \frac{1}{4} + b + \frac{3}{4} = 1 + b - a \Rightarrow b = a \Rightarrow

F (x) = \frac{a}{x - 1} + \frac{1}{4 {(x - 1)}^{2}} + \frac{a}{x + 1} + \frac{3}{4 {(x + 1)}^{2}}

= \frac{2 a x}{x^{2} - 1} + \frac{1}{4 {(x - 1)}^{2}} + \frac{3}{4 {(x + 1)}^{2}}

{l i m}_{x \to + \infty} x F (x) = 0 = 2 a \Rightarrow a = 0 \Rightarrow

F (x) = \frac{1}{4 {(x - 1)}^{2}} + \frac{3}{4 {(x + 1)}^{2}} \Rightarrow

S_{n} = \frac{1}{4} \sum_{k = 2}^{n} \frac{1}{{(k - 1)}^{2}} + \frac{3}{4} \sum_{k = 2}^{n} \frac{1}{{(k + 1)}^{2}} b u t

\sum_{k = 2}^{n} \frac{1}{{(k - 1)}^{2}} = \sum_{k = 1}^{n - 1} \frac{1}{k^{2}} = ξ_{n - 1} (2) \to ξ (2) = \frac{π^{2}}{6}

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

= ξ_{n + 1} (2) - \frac{3}{4} \to ξ (2) - \frac{3}{4} = \frac{π^{2}}{6} - \frac{3}{4}

{l i m}_{n \to + \infty} S_{n} = \frac{1}{4} \frac{π^{2}}{6} + \frac{3}{4} (\frac{π^{2}}{6} - \frac{3}{4})

= \frac{π^{2}}{24} + \frac{3 π^{2}}{24} - \frac{9}{16} = \frac{π^{2}}{6} - \frac{9}{16} .

Commented by sma3l2996 last updated on 26/Apr/18

I t h i n k y o u h a v e a n e r r o r o n l i n e s 14 a n d 15

- 1 - \frac{1}{4} = \frac{- 5}{4}

Commented by prof Abdo imad last updated on 26/Apr/18

y e s y e s y o u a r e r i g h t i w i l l c o r r e c t t h e a n s w e r

t h a n k y o u s o m u c h s i r s m a 3 l \dots .

Commented by prof Abdo imad last updated on 26/Apr/18

e r r o r f r o m l i n e 15 \sum_{k = 2}^{n} \frac{1}{{(k + 1)}^{2}} = ξ_{n + 1} (2) - \frac{5}{4}

\to_{n \to + \infty} ξ (2) - \frac{5}{4} = \frac{π^{2}}{6} - \frac{5}{4} \Rightarrow

{l i m}_{n \to + \infty} S_{n} = \frac{1}{4} \frac{π^{2}}{6} + \frac{3}{4} (\frac{π^{2}}{6} - \frac{5}{4})

= \frac{π^{2}}{24} + \frac{3 π^{2}}{24} - \frac{15}{16} = \frac{π^{2}}{6} - \frac{15}{16}

★ l i m S_{n} = \frac{π^{2}}{6} - \frac{15}{16} ★

Answered by sma3l2996 last updated on 23/Apr/18

A_{n} = \underset{n = 2}{\sum^{\infty}} \frac{n^{2} - n + 1}{{(n - 1)}^{2} {(n + 1)}^{2}} = \underset{n = 2}{\sum^{\infty}} (\frac{a}{n - 1} + \frac{b}{{(n - 1)}^{2}} + \frac{c}{n + 1} + \frac{d}{{(n + 1)}^{2}})

w i t h a = c = 0; b = \frac{1}{4}; d = \frac{3}{4}

\underset{n = 2}{\sum^{\infty}} \frac{n^{2} - n + 1}{{(n - 1)}^{2} {(n + 1)}^{2}} = \frac{1}{4} \underset{n = 2}{\sum^{\infty}} (\frac{1}{{(n - 1)}^{2}} + \frac{3}{{(n + 1)}^{2}})

= \frac{1}{4} \underset{n = 2}{\sum^{\infty}} \frac{1}{{(n - 1)}^{2}} + \frac{3}{4} \underset{n = 2}{\sum^{\infty}} \frac{1}{{(n + 1)}^{2}}

l e t m = n - 1 a n d k = n + 1

s o \underset{n = 2}{\sum^{\infty}} \frac{n^{2} - n + 1}{{(n - 1)}^{2} {(n + 1)}^{2}} = \frac{1}{4} \underset{m = 1}{\sum^{\infty}} \frac{1}{m^{2}} + \frac{3}{4} \underset{k = 3}{\sum^{\infty}} \frac{1}{k^{2}}

= \frac{1}{4} ξ (2) + \frac{3}{4} (ξ (2) - (\frac{1}{1^{2}} + \frac{1}{2^{2}})) = ξ (2) - \frac{5}{4} w i t h (ξ (x) i s z e t a f u n c t i o n)

= \frac{π^{2}}{6} - \frac{5}{4}

Commented by prof Abdo imad last updated on 26/Apr/18

y o u h a v e c o m m i t e d a e r r o r o f c a l c u l u s s i r s m a 3 l . .

Commented by sma3l2996 last updated on 26/Apr/18

\frac{1}{4} ξ (2) + \frac{3}{4} (ξ (2) - \frac{5}{4}) = ξ (2) - \frac{15}{16}

= \frac{π^{2}}{6} - \frac{15}{16}