calculate Σ_(n=1) ^∞ (1/(n(n+1)(n+2)(n+3)(n+4)(n+5)))


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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
$let decompose F(x)=(1/(x(x+1)(x+2)(x+3)(x+4)(x+5))) F(x)=(a/(x )) +(b/(x+1)) +(c/(x+2)) +(d/(x+3)) +(e/(x+4)) +(f/(x+5)) a =lim_(x→0) xF(x)=(1/(1.2.3.4.5)) =(1/(120)) b=lim_(x→−1) (x+1)F(x)=(1/((−1)1.2.3.4)) =((−1)/(24)) c=lim_(x→−2) (x+2)F(x)= (1/((−2).(−1)1.2.3)) =(1/(12)) d =lim_(x→−3) (x+3)F(x)=(1/((−3)(−2)(−1)1.2)) =((−1)/(12)) e =lim_(x→−4) (x+4)F(x)=(1/((−4)(−3)(−2)(−1).1)) =(1/(24)) f =lim_(x→−5) (x+5)F(x) =(1/((−5)(−4)(−3)(−2)(−1))) =((−1)/(120)) ⇒ F(x)=(1/(120x)) −(1/(24(x+1))) +(1/(12(x+2))) −(1/(12(x+3))) +(1/(24(x+4))) −(1/(120(x+5))) ⇒ Σ_(k=1) ^n F(k) =(1/(120)) Σ_(k=1) ^n (1/k) −(1/(24)) Σ_(k=1) ^n (1/(k+1)) +(1/(12))Σ_(k=1) ^n (1/(k+2)) −(1/(12))Σ_(k=1) ^n (1/(k+3)) +(1/(24)) Σ_(k=1) ^n (1/(k+4)) −(1/(120))Σ_(k=1) ^n (1/(k+5)) =(1/(120)) H_n −(1/(24)){ H_(n+1) −1} +(1/(12)){H_(n+2) −(3/2)} −(1/(12)){H_(n+3) −(3/2)−(1/3)} +(1/(24)){H_(n+4) −(3/2)−(1/3)−(1/4)}−(1/(120)){H_(n+5) −(3/2)−(1/3)−(1/4) −(1/5)} =(1/(120)){H_n −H_(n+5) } +(1/(24)){H_(n+4) −H_(n+1) } +(1/(12)){ H_(n+2) −H_(n+3) } +(1/(24)) −(1/8) +(1/(12))((3/2)+(1/3))−(1/(24))((3/2)+(1/3)+(1/4))+(1/(120)){(3/2) +(1/3) +(1/4) +(1/5)} but H_n −H_(n+5) →0 ,H_(n+4) −H_(n+1) →0 ^� H_(n+2) −H_(n+3) →0 (n→+∞) after reducton of calculus we get lim_(n→+∞) Σ_(k=1) ^n F(k) =(1/(600)) =S .$
$l e t d e c o m p o s e F (x) = \frac{1}{x (x + 1) (x + 2) (x + 3) (x + 4) (x + 5)}$ $F (x) = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{x + 2} + \frac{d}{x + 3} + \frac{e}{x + 4} + \frac{f}{x + 5}$ $a = {l i m}_{x \to 0} x F (x) = \frac{1}{1.2.3.4.5} = \frac{1}{120}$ $b = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{1}{(- 1) 1.2.3.4} = \frac{- 1}{24}$ $c = {l i m}_{x \to - 2} (x + 2) F (x) = \frac{1}{(- 2) . (- 1) 1.2.3} = \frac{1}{12}$ $d = {l i m}_{x \to - 3} (x + 3) F (x) = \frac{1}{(- 3) (- 2) (- 1) 1.2} = \frac{- 1}{12}$ $e = {l i m}_{x \to - 4} (x + 4) F (x) = \frac{1}{(- 4) (- 3) (- 2) (- 1) . 1} = \frac{1}{24}$ $f = {l i m}_{x \to - 5} (x + 5) F (x) = \frac{1}{(- 5) (- 4) (- 3) (- 2) (- 1)} = \frac{- 1}{120} \Rightarrow$ $F (x) = \frac{1}{120 x} - \frac{1}{24 (x + 1)} + \frac{1}{12 (x + 2)} - \frac{1}{12 (x + 3)} + \frac{1}{24 (x + 4)} - \frac{1}{120 (x + 5)} \Rightarrow$ $\sum_{k = 1}^{n} F (k) = \frac{1}{120} \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{24} \sum_{k = 1}^{n} \frac{1}{k + 1} + \frac{1}{12} \sum_{k = 1}^{n} \frac{1}{k + 2} - \frac{1}{12} \sum_{k = 1}^{n} \frac{1}{k + 3}$ $+ \frac{1}{24} \sum_{k = 1}^{n} \frac{1}{k + 4} - \frac{1}{120} \sum_{k = 1}^{n} \frac{1}{k + 5}$ $= \frac{1}{120} H_{n} - \frac{1}{24} {H_{n + 1} - 1} + \frac{1}{12} {H_{n + 2} - \frac{3}{2}} - \frac{1}{12} {H_{n + 3} - \frac{3}{2} - \frac{1}{3}}$ $+ \frac{1}{24} {H_{n + 4} - \frac{3}{2} - \frac{1}{3} - \frac{1}{4}} - \frac{1}{120} {H_{n + 5} - \frac{3}{2} - \frac{1}{3} - \frac{1}{4} - \frac{1}{5}}$ $= \frac{1}{120} {H_{n} - H_{n + 5}} + \frac{1}{24} {H_{n + 4} - H_{n + 1}} + \frac{1}{12} {H_{n + 2} - H_{n + 3}}$ $+ \frac{1}{24} - \frac{1}{8} + \frac{1}{12} (\frac{3}{2} + \frac{1}{3}) - \frac{1}{24} (\frac{3}{2} + \frac{1}{3} + \frac{1}{4}) + \frac{1}{120} {\frac{3}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}} b u t$ $H_{n} - H_{n + 5} \to 0, H_{n + 4} - H_{n + 1} \to 0 \bar{} H_{n + 2} - H_{n + 3} \to 0 (n \to + \infty)$ $a f t e r r e d u c t o n o f c a l c u l u s w e g e t$ ${l i m}_{n \to + \infty} \sum_{k = 1}^{n} F (k) = \frac{1}{600} = S .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

	$f i r s t c a l c u l a t e S_{n} t h e n t h e v a l u e w h e n n \to \infty$ $S_{n} = C - \frac{1}{(n + 1) (n + 2) (n + 3) (n + 4) (n + 5) \times 1 \times 5}$ $S_{1} = T_{1} = \frac{1}{1 \times 2 \times 3 \times 4 \times 5 \times 6} = \frac{1}{720}$ $s o S_{n} = C - \frac{1}{(n + 1) (n + 2) (n + 3) (n + 4) (n + 5) \times 1 \times 5}$ $\frac{1}{720} = C - \frac{1}{2 \times 3 \times 4 \times 5 \times 6 \times 1 \times 5}$ $C = \frac{1}{720} + \frac{1}{720 \times 5} = \frac{1}{720} (1 + \frac{1}{5}) = \frac{1}{720} \times \frac{6}{5}$ $C = \frac{1}{600}$ $s 0 S_{n} = \frac{1}{600} - \frac{1}{(n + 1) (n + 2) (n + 3) (n + 4) (n + 5) \times 1 \times 5}$ $w h e n n \to \infty S_{n} \to \frac{1}{600}$ $s o r e a u i r e d a n s w e r i s \frac{1}{600}$
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