calculate-n-1-1-n-n-1-n-2-n-3-n-4-n-5-

Question Number 46617 by maxmathsup by imad last updated on 29/Oct/18

c a l c u l a t e \sum_{n = 1}^{\infty} \frac{1}{n (n + 1) (n + 2) (n + 3) (n + 4) (n + 5)}

Commented by maxmathsup by imad last updated on 29/Oct/18

l e t d e c o m p o s e F (x) = \frac{1}{x (x + 1) (x + 2) (x + 3) (x + 4) (x + 5)}

F (x) = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{x + 2} + \frac{d}{x + 3} + \frac{e}{x + 4} + \frac{f}{x + 5}

a = {l i m}_{x \to 0} x F (x) = \frac{1}{1.2.3.4.5} = \frac{1}{120}

b = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{1}{(- 1) 1.2.3.4} = \frac{- 1}{24}

c = {l i m}_{x \to - 2} (x + 2) F (x) = \frac{1}{(- 2) . (- 1) 1.2.3} = \frac{1}{12}

d = {l i m}_{x \to - 3} (x + 3) F (x) = \frac{1}{(- 3) (- 2) (- 1) 1.2} = \frac{- 1}{12}

e = {l i m}_{x \to - 4} (x + 4) F (x) = \frac{1}{(- 4) (- 3) (- 2) (- 1) . 1} = \frac{1}{24}

f = {l i m}_{x \to - 5} (x + 5) F (x) = \frac{1}{(- 5) (- 4) (- 3) (- 2) (- 1)} = \frac{- 1}{120} \Rightarrow

F (x) = \frac{1}{120 x} - \frac{1}{24 (x + 1)} + \frac{1}{12 (x + 2)} - \frac{1}{12 (x + 3)} + \frac{1}{24 (x + 4)} - \frac{1}{120 (x + 5)} \Rightarrow

\sum_{k = 1}^{n} F (k) = \frac{1}{120} \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{24} \sum_{k = 1}^{n} \frac{1}{k + 1} + \frac{1}{12} \sum_{k = 1}^{n} \frac{1}{k + 2} - \frac{1}{12} \sum_{k = 1}^{n} \frac{1}{k + 3}

+ \frac{1}{24} \sum_{k = 1}^{n} \frac{1}{k + 4} - \frac{1}{120} \sum_{k = 1}^{n} \frac{1}{k + 5}

= \frac{1}{120} H_{n} - \frac{1}{24} {H_{n + 1} - 1} + \frac{1}{12} {H_{n + 2} - \frac{3}{2}} - \frac{1}{12} {H_{n + 3} - \frac{3}{2} - \frac{1}{3}}

+ \frac{1}{24} {H_{n + 4} - \frac{3}{2} - \frac{1}{3} - \frac{1}{4}} - \frac{1}{120} {H_{n + 5} - \frac{3}{2} - \frac{1}{3} - \frac{1}{4} - \frac{1}{5}}

= \frac{1}{120} {H_{n} - H_{n + 5}} + \frac{1}{24} {H_{n + 4} - H_{n + 1}} + \frac{1}{12} {H_{n + 2} - H_{n + 3}}

+ \frac{1}{24} - \frac{1}{8} + \frac{1}{12} (\frac{3}{2} + \frac{1}{3}) - \frac{1}{24} (\frac{3}{2} + \frac{1}{3} + \frac{1}{4}) + \frac{1}{120} {\frac{3}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}} b u t

H_{n} - H_{n + 5} \to 0, H_{n + 4} - H_{n + 1} \to 0 \bar{} H_{n + 2} - H_{n + 3} \to 0 (n \to + \infty)

a f t e r r e d u c t o n o f c a l c u l u s w e g e t

{l i m}_{n \to + \infty} \sum_{k = 1}^{n} F (k) = \frac{1}{600} = S .

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

f i r s t c a l c u l a t e S_{n} t h e n t h e v a l u e w h e n n \to \infty

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

S_{1} = T_{1} = \frac{1}{1 \times 2 \times 3 \times 4 \times 5 \times 6} = \frac{1}{720}

s o S_{n} = C - \frac{1}{(n + 1) (n + 2) (n + 3) (n + 4) (n + 5) \times 1 \times 5}

\frac{1}{720} = C - \frac{1}{2 \times 3 \times 4 \times 5 \times 6 \times 1 \times 5}

C = \frac{1}{720} + \frac{1}{720 \times 5} = \frac{1}{720} (1 + \frac{1}{5}) = \frac{1}{720} \times \frac{6}{5}

C = \frac{1}{600}

s 0 S_{n} = \frac{1}{600} - \frac{1}{(n + 1) (n + 2) (n + 3) (n + 4) (n + 5) \times 1 \times 5}

w h e n n \to \infty S_{n} \to \frac{1}{600}

s o r e a u i r e d a n s w e r i s \frac{1}{600}