find the value of Σ_(n=2) ^∞ (n/((n+1)^2 (n−1)^3 ))


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Question Number 65674 by mathmax by abdo last updated on 01/Aug/19

$f i n d t h e v a l u e o f \sum_{n = 2}^{\infty} \frac{n}{{(n + 1)}^{2} {(n - 1)}^{3}}$
Commented by mathmax by abdo last updated on 04/Aug/19
$let S =Σ_(n=2) ^∞ (n/((n+1)^2 (n−1)^3 )) ⇒S =Σ_(n=1) ^∞ ((n+1)/((n+2)^2 n^3 )) let decompose F(x) =((x+1)/(x^3 (x+2)^2 )) ⇒ F(x) =(a/x) +(b/x^2 ) +(c/x^3 ) +(d/(x+2)) +(e/((x+2)^2 )) c =lim_(x→0) x^3 F(x) =(1/4) e =lim_(x→−2) (x+2)^2 F(x) =((−1)/(−8))=(1/8) ⇒ F(x) =(a/x) +(b/x^2 ) +(1/(4x^3 )) +(d/(x+2)) +(1/(8(x+2)^2 )) lim_(x→+∞) xF(x) =0 =a+d ⇒d =−a ⇒ F(x) =(a/x) +(b/x^2 ) +(1/(4x^3 ))−(a/(x+2)) +(1/(8(x+2)^2 )) F(1) =(2/9) =a+b+(1/4)−(a/3) +(1/(72)) =(2/3)a +b +(1/(72)) ⇒ 2 =6a+9b +(1/8) ⇒6a+9b =2−(1/8) =((15)/8) F(−1) =0 =−a+b−(1/4)−a +(1/8) =−2a+b−(1/8) ⇒ −2a+b =(1/8) ⇒b =2a+(1/8) ⇒6a+9(2a+(1/8))=((15)/8) ⇒ 24a +(9/8) =((15)/8) ⇒24a=(6/8) =(3/4) ⇒a =(3/(4.8.3)) =(1/(32)) b =(1/(16))+(1/8) =(3/(16)) ⇒F(x) =(1/(32x)) +(3/(16x^2 )) +(1/(4x^3 )) −(1/(32(x+2))) +(1/(8(x+2)^2 )) S_n =Σ_(k=1) ^n F(k) =(1/(32))Σ_(k=1) ^n (1/k) +(3/(16)) Σ_(k=1) ^n (1/k^2 ) +(1/4)Σ_(k=1) ^n (1/k^3 ) −(1/(32))Σ_(k=1) ^n (1/(k+2)) +(1/8)Σ_(k=1) ^n (1/((x+2)^2 )) Σ_(k=1) ^n (1/(k+2)) =Σ_(k=3) ^(n+2) (1/k) =Σ_(k=1) ^n (1/k)−(3/2) +(1/(n+1)) +(1/(n+2)) ⇒ (1/(32))Σ_(k=1) ^n (1/k) −(1/(32))Σ_(k=1) ^n (1/(k+2)) =−(1/(32))(−(3/2) +(1/(n+1)) +(1/(n+2)))→(3/(64)) ⇒ Σ_(k=1) ^n (1/((x+2)^2 )) =Σ_(k=3) ^(n+2) (1/k^2 ) =ξ_n (2)+(1/((n+1)^2 )) +(1/((n+2)^2 ))−1−(1/4) →ξ(2)−(5/4) ⇒ S =lim_(n→+∞) S_n =(3/(64)) +(3/(16))ξ(2) +(1/4)ξ(3) +(1/8){ξ(2)−(5/4)} =(3/(64)) +(5/(16)) (π^2 /6) +(1/4)ξ(3)−(5/(32)) =−(7/(64)) +((5π^2 )/(96)) +(1/4)ξ(3)$
$l e t S = \sum_{n = 2}^{\infty} \frac{n}{{(n + 1)}^{2} {(n - 1)}^{3}} \Rightarrow S = \sum_{n = 1}^{\infty} \frac{n + 1}{{(n + 2)}^{2} n^{3}}$ $l e t d e c o m p o s e F (x) = \frac{x + 1}{x^{3} {(x + 2)}^{2}} \Rightarrow$ $F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x^{3}} + \frac{d}{x + 2} + \frac{e}{{(x + 2)}^{2}}$ $c = {l i m}_{x \to 0} x^{3} F (x) = \frac{1}{4}$ $e = {l i m}_{x \to - 2} {(x + 2)}^{2} F (x) = \frac{- 1}{- 8} = \frac{1}{8} \Rightarrow$ $F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{1}{4 x^{3}} + \frac{d}{x + 2} + \frac{1}{8 {(x + 2)}^{2}}$ ${l i m}_{x \to + \infty} x F (x) = 0 = a + d \Rightarrow d = - a \Rightarrow$ $F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{1}{4 x^{3}} - \frac{a}{x + 2} + \frac{1}{8 {(x + 2)}^{2}}$ $F (1) = \frac{2}{9} = a + b + \frac{1}{4} - \frac{a}{3} + \frac{1}{72} = \frac{2}{3} a + b + \frac{1}{72} \Rightarrow$ $2 = 6 a + 9 b + \frac{1}{8} \Rightarrow 6 a + 9 b = 2 - \frac{1}{8} = \frac{15}{8}$ $F (- 1) = 0 = - a + b - \frac{1}{4} - a + \frac{1}{8} = - 2 a + b - \frac{1}{8} \Rightarrow$ $- 2 a + b = \frac{1}{8} \Rightarrow b = 2 a + \frac{1}{8} \Rightarrow 6 a + 9 (2 a + \frac{1}{8}) = \frac{15}{8} \Rightarrow$ $24 a + \frac{9}{8} = \frac{15}{8} \Rightarrow 24 a = \frac{6}{8} = \frac{3}{4} \Rightarrow a = \frac{3}{4.8.3} = \frac{1}{32}$ $b = \frac{1}{16} + \frac{1}{8} = \frac{3}{16} \Rightarrow F (x) = \frac{1}{32 x} + \frac{3}{16 x^{2}} + \frac{1}{4 x^{3}} - \frac{1}{32 (x + 2)} + \frac{1}{8 {(x + 2)}^{2}}$ $S_{n} = \sum_{k = 1}^{n} F (k) = \frac{1}{32} \sum_{k = 1}^{n} \frac{1}{k} + \frac{3}{16} \sum_{k = 1}^{n} \frac{1}{k^{2}} + \frac{1}{4} \sum_{k = 1}^{n} \frac{1}{k^{3}}$ $- \frac{1}{32} \sum_{k = 1}^{n} \frac{1}{k + 2} + \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{{(x + 2)}^{2}}$ $\sum_{k = 1}^{n} \frac{1}{k + 2} = \sum_{k = 3}^{n + 2} \frac{1}{k} = \sum_{k = 1}^{n} \frac{1}{k} - \frac{3}{2} + \frac{1}{n + 1} + \frac{1}{n + 2} \Rightarrow$ $\frac{1}{32} \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{32} \sum_{k = 1}^{n} \frac{1}{k + 2} = - \frac{1}{32} (- \frac{3}{2} + \frac{1}{n + 1} + \frac{1}{n + 2}) \to \frac{3}{64} \Rightarrow$ $\sum_{k = 1}^{n} \frac{1}{{(x + 2)}^{2}} = \sum_{k = 3}^{n + 2} \frac{1}{k^{2}} = ξ_{n} (2) + \frac{1}{{(n + 1)}^{2}} + \frac{1}{{(n + 2)}^{2}} - 1 - \frac{1}{4}$ $\to ξ (2) - \frac{5}{4} \Rightarrow S = {l i m}_{n \to + \infty} S_{n}$ $= \frac{3}{64} + \frac{3}{16} ξ (2) + \frac{1}{4} ξ (3) + \frac{1}{8} {ξ (2) - \frac{5}{4}}$ $= \frac{3}{64} + \frac{5}{16} \frac{π^{2}}{6} + \frac{1}{4} ξ (3) - \frac{5}{32} = - \frac{7}{64} + \frac{5 π^{2}}{96} + \frac{1}{4} ξ (3)$
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