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

Question Number 33588 by abdo imad last updated on 19/Apr/18

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

Commented by abdo imad last updated on 20/Apr/18

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

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

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

b = {l i m}_{x \to 0} x^{2} F (x) = 1

c = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{1}{- 1} = - 1

d = {l i m}_{x \to - \frac{1}{2}} (2 x + 1) F (x) = \frac{1}{\frac{1}{4} . \frac{1}{2}} = 8 \Rightarrow

F (x) = \frac{a}{x} + \frac{1}{x^{2}} - \frac{1}{x + 1} + \frac{8}{2 x + 1}

F (1) = \frac{1}{6} = a + 1 - \frac{1}{2} + \frac{8}{3} = a + \frac{1}{2} + \frac{8}{3} \Rightarrow

1 = 6 a + 3 + 16 = 6 a + 19 \Rightarrow 6 a = - 18 \Rightarrow a = - 3 \Rightarrow

F (x) = - \frac{3}{x} + \frac{1}{x^{2}} - \frac{1}{x + 1} + \frac{8}{2 x + 1}

S_{n} = - 3 \sum_{k = 1}^{n} \frac{1}{k} + \sum_{k = 1}^{n} \frac{1}{k^{2}} - \sum_{k = 1}^{n} \frac{1}{k + 1} + 8 \sum_{k = 1}^{n} \frac{1}{2 k + 1}

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

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

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

\sum_{k = 1}^{n} \frac{1}{2 k + 1} = \frac{1}{3} + \frac{1}{5} + \dots \dots + \frac{1}{2 n + 1}

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

= H_{2 n + 1} - \frac{1}{2} H_{n} - 1 \Rightarrow

S_{n} = - 3 H_{n} + ξ_{n} (2) - H_{n + 1} + 1 + 8 H_{2 n + 1} - 4 H_{n} - 8

S_{n} = - 7 H_{n} - H_{n + 1} + 8 H_{2 n + 1} - 7 + ξ_{n} (2) \Rightarrow

S_{n} = - 7 (l n (n) + γ + o (\frac{1}{n})) - (l n (n + 1) + γ + o (\frac{1}{n}))

+ 8 (l n (2 n + 1) + γ + o (\frac{1}{n})) + ξ_{n} (2) - 7

= - l n (n^{7} (n + 1)) + l n ({(2 n + 1)}^{8}) + ξ_{n} (2) - 7

= l n (\frac{{(2 n + 1)}^{8}}{n^{8} + n^{7}}) + ξ_{n} (2) - 7 \Rightarrow

{l i m}_{n \to + \infty} S_{n} = l n (2^{8}) + ξ (2) - 7 = 8 l n (2) + \frac{π^{2}}{6} - 7 . s o

\sum_{n = 1}^{\infty} \frac{1}{n^{2} (n + 1) (2 n + 1)} = 8 l n (2) + \frac{π^{2}}{6} - 7 .

Answered by sma3l2996 last updated on 19/Apr/18

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

w i t h a = - 3; b = 1; c = - 1; d = 8

s o \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (n + 1) (2 n + 1)} = \underset{n = 1}{\sum^{\infty}} (\frac{- 3}{n} + \frac{1}{n^{2}} - \frac{1}{n + 1} + \frac{8}{2 n + 1})

{l e t}^{'} s c a l c u l a t e A = \underset{n = 1}{\sum^{\infty}} \frac{8}{2 n + 1}

A = 8 (\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots) = 8 [(\frac{1}{3} + \frac{1}{5} + \dots) + (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots) - (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots)]

= 8 [\underset{n = 2}{\sum^{\infty}} \frac{1}{n} - \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots)]

= 8 (\underset{n = 2}{\sum^{\infty}} \frac{1}{n} - \frac{1}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{n} - \frac{1}{2}) = 4 \underset{n = 2}{\sum^{\infty}} \frac{1}{n} - 4

s o \underset{n = 1}{\sum^{\infty}} (\frac{8}{2 n + 1} - \frac{3}{n} - \frac{1}{n + 1}) = 4 \underset{n = 2}{\sum^{\infty}} \frac{1}{n} - 4 - 3 (\underset{n = 2}{\sum^{\infty}} \frac{1}{n} + \frac{1}{1}) - \underset{n = 1}{\sum^{\infty}} \frac{1}{n + 1}

l e t k = n + 1

\underset{n = 1}{\sum^{\infty}} (\frac{8}{2 n + 1} - \frac{3}{n} - \frac{1}{n + 1}) = \underset{n = 2}{\sum^{\infty}} \frac{1}{n} - 7 - \underset{k = 2}{\sum^{\infty}} \frac{1}{k} = - 7

w e k n o w t h a t ζ (2) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6} (ζ (x) i s R i e m a n n z e t a f u n c t i o n)

A s r e s u l t s

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (n + 1) (2 n + 1)} = \frac{π^{2}}{6} - 7

Commented by abdo imad last updated on 20/Apr/18

s i r s m a 3 l 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 b e c a u s e

t h e s e r i e i s p o s i t i f b u t t h e v a l u e \frac{π^{2}}{6} - 7 i s n e g a t i v e b u t

n e v e r m i n d y o u r m e t h o d i s c o r r e c t \dots

Commented by sma3l2996 last updated on 20/Apr/18

T h a t w h a t I j u s t s a w . T h e r e i s a n e r r o r i n m y w o r k .

C a n y o u c h e c k m y w o r k p l e a s e ?

Commented by abdo imad last updated on 20/Apr/18

i a d v i s e y o u s i r s m a 3 l t o u s e t h e h a r m o n i c s e q u e n c e H_{n}

t o o k a l o o k i n m y m e t h o d \dots .

Commented by sma3l2996 last updated on 20/Apr/18

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