find-the-value-of-n-1-1-n-1-n-2-n-3-in-terms-of-3-we-give-x-n-1-1-n-x-and-x-gt-1-zeta-function-of-Rieman-

Question Number 26222 by abdo imad last updated on 22/Dec/17

f i n d t h e v a l u e o f \sum_{n = 1}^{\propto} \frac{1}{(n + 1) (n + 2) n^{3}} i n t e r m s o f ξ (3)

w e g i v e ξ (x) = \sum_{n = 1}^{\propto} \frac{1}{n^{x}} a n d x > 1

(z e t a f u n c t i o n o f R i e m a n)

Commented by abdo imad last updated on 24/Dec/17

w e d e c o m p o s e t h e f r a c t i o n F (x) = \frac{1}{(x + 1) (x + 2) x^{3}} i n s i m p l e

e l e m e n t s F (x) = \frac{a}{x + 1} + \frac{b}{x + 2} + \frac{c}{x} + \frac{d}{x^{2}} + \frac{e}{x^{3}}

a = {l i m}_{x - > - 1} (x + 1) F (x) = - 1

b = {l i m}_{x - > - 2} (x + 2) F (x) = \frac{1}{8}

e = {l i m}_{x - > 0} x^{3} F (x) = \frac{1}{2} \Rightarrow F (x) = - \frac{1}{x + 1} + \frac{1}{8 (x + 2)} + \frac{c}{x} + \frac{d}{x^{2}} + \frac{1}{2 x^{3}}

{l i m}_{x - >\propto} x F (x) = 0 = - 1 + \frac{1}{8} + c \Rightarrow c = \frac{7}{8}

F (x) = - \frac{1}{x + 1} + \frac{1}{8 (x + 2)} + \frac{7}{8 x} + \frac{d}{x^{2}} + \frac{1}{2 x^{3}}

F (1) = \frac{1}{6} = - \frac{1}{2} + \frac{1}{24} + \frac{7}{8} + d + \frac{1}{2} = \frac{11}{12} + d \Rightarrow d = \frac{1}{6} - \frac{11}{12} = - \frac{3}{4}

F (x) = - \frac{1}{x + 1} + \frac{1}{8 (x + 2)} + \frac{7}{8 x} - \frac{3}{4 x^{2}} + \frac{1}{2 x^{3}}

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

S_{n} = - \sum_{k = 1}^{k = n} \frac{1}{k + 1} + \frac{1}{8} \sum_{k = 1}^{k = n} \frac{1}{k + 2} + \frac{7}{8} \sum_{k = 1}^{k = n} \frac{1}{k} - \frac{3}{4} \sum_{k = 1}^{k = n} \frac{1}{k^{2}} + \frac{1}{2} \sum_{k = 1}^{k = n} \frac{1}{n^{3}}

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

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

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

S_{n} = - H_{n + 1} + 1 + \frac{1}{8} H_{n + 2} - \frac{3}{16} + \frac{7}{8} H_{n} - \frac{3}{4} ξ_{n} (2) + \frac{1}{2} ξ_{n} (3)

= \frac{13}{16} + \frac{1}{8} H_{n + 2} + \frac{7}{8} H_{n} - H_{n + 1} - \frac{3}{4} ξ_{n} (2) + \frac{1}{2} ξ_{n} (3)

b u t \frac{1}{8} H_{n + 2} + \frac{7}{8} H_{n} - H_{n + 1} = \frac{1}{8} (l n (n + 2) + γ + o (\frac{1}{n})) + \frac{7}{8} (l n (n) + γ + o (\frac{1}{n}) - (l n (n + 1) + γ + o (\frac{1}{n}))

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

= l n (\frac{{(n + 2)}^{\frac{1}{8}} . n^{\frac{7}{8}}}{n + 1}) - - - > 0_{n - >\propto}

{l i m}_{n - >\propto} ξ_{n} (2) = ξ (2) a n d {l i m}_{n - >\propto} ξ_{n} (3) = ξ (3) \Rightarrow

{l i m}_{n - >\propto} S_{n} = \frac{13}{16} + \frac{1}{2} ξ (3) - \frac{3}{4} ξ (2)

Commented by abdo imad last updated on 25/Dec/17

ξ_{n} (x) = \sum_{k = 1}^{k = n} \frac{1}{k^{x}} w i t h x > 1