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


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Question Number 63665 by mathmax by abdo last updated on 07/Jul/19

$f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} {(n + 1)}^{3}}$
Commented by mathmax by abdo last updated on 07/Jul/19
$let decompose F(x)=(1/(x^2 (x+1)^3 )) ⇒ F(x)=(a/x) +(b/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(e/((x+1)^3 )) b=lim_(x→0) x^2 F(x)=1 e =lim_(x→−1) (x+1)^3 F(x)=1 ⇒F(x)=(a/x) +(1/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) F(x)−(1/x^2 ) =(1/x^2 ){(1/((x+1)^3 )) −1} =(1/x^2 ){((1−x^3 −3x^2 −3x−1)/((x+1)^3 ))} =−((x^3 +3x^2 +3x)/(x^2 (x+1)^3 )) =−((x^2 +3x +3)/(x(x+1)^3 )) =(a/x) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) ⇒ −((x^2 +3x +3)/((x+1)^3 )) =a +((cx)/(x+1)) +(dx/((x+1)^2 )) +(x/((x+1)^3 )) (x→0) ⇒ a =−3 ⇒F(x) =−(3/x) +(1/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) lim_(x→+∞) xF(x)=0 =−3 +c ⇒c=3 ⇒ F(x)=−(3/x) +(1/x^2 ) +(3/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) F(−2) =−(1/4) =(3/2) +(1/4) −3 +d −1 =(7/4)−4 +d ⇒ d =−(1/4)−(7/4) +4 =−2+4 =2 ⇒ F(x)=((−3)/x) +(1/x^2 ) +(3/(x+1)) +(2/((x+1)^2 )) +(1/((x+1)^3 )) S =Σ_(n=1) ^∞ (−1)^n F(n) =−3 Σ_(n=1) ^∞ (((−1)^n )/n) +Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +2Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) +Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) we have Σ_(n=1) ^∞ (x^n /n) =−ln(1−x) if ∣x∣<1 ⇒Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) if δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) we have proved that δ(x)=(2^(1−x) −1)ξ(x) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(2^(−1) −1)ξ(2) =−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^2 ) =−{−(π^2 /(12)) +1}=(π^2 /(12)) −1 also Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^3 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^3 ) =−{ δ(3) +1} =−(2^(−2) −1)ξ(3)−1 =−(−(3/4))ξ(3)−1 =(3/4)ξ(3)−1 ⇒ S = 3ln(2)−(π^2 /(12)) +((2π^2 )/(12)) −2 +(3/4)ξ(3)−1 =3ln(2) +(π^2 /(12)) +(3/4)ξ(3)−3 .$
$l e t d e c o m p o s e F (x) = \frac{1}{x^{2} {(x + 1)}^{3}} \Rightarrow$ $F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x + 1} + \frac{d}{{(x + 1)}^{2}} + \frac{e}{{(x + 1)}^{3}}$ $b = {l i m}_{x \to 0} x^{2} F (x) = 1$ $e = {l i m}_{x \to - 1} {(x + 1)}^{3} F (x) = 1 \Rightarrow F (x) = \frac{a}{x} + \frac{1}{x^{2}} + \frac{c}{x + 1} + \frac{d}{{(x + 1)}^{2}} + \frac{1}{{(x + 1)}^{3}}$ $F (x) - \frac{1}{x^{2}} = \frac{1}{x^{2}} {\frac{1}{{(x + 1)}^{3}} - 1} = \frac{1}{x^{2}} {\frac{1 - x^{3} - 3 x^{2} - 3 x - 1}{{(x + 1)}^{3}}}$ $= - \frac{x^{3} + 3 x^{2} + 3 x}{x^{2} {(x + 1)}^{3}} = - \frac{x^{2} + 3 x + 3}{x {(x + 1)}^{3}} = \frac{a}{x} + \frac{c}{x + 1} + \frac{d}{{(x + 1)}^{2}} + \frac{1}{{(x + 1)}^{3}} \Rightarrow$ $- \frac{x^{2} + 3 x + 3}{{(x + 1)}^{3}} = a + \frac{c x}{x + 1} + \frac{d x}{{(x + 1)}^{2}} + \frac{x}{{(x + 1)}^{3}} (x \to 0) \Rightarrow$ $a = - 3 \Rightarrow F (x) = - \frac{3}{x} + \frac{1}{x^{2}} + \frac{c}{x + 1} + \frac{d}{{(x + 1)}^{2}} + \frac{1}{{(x + 1)}^{3}}$ ${l i m}_{x \to + \infty} x F (x) = 0 = - 3 + c \Rightarrow c = 3 \Rightarrow$ $F (x) = - \frac{3}{x} + \frac{1}{x^{2}} + \frac{3}{x + 1} + \frac{d}{{(x + 1)}^{2}} + \frac{1}{{(x + 1)}^{3}}$ $F (- 2) = - \frac{1}{4} = \frac{3}{2} + \frac{1}{4} - 3 + d - 1 = \frac{7}{4} - 4 + d \Rightarrow$ $d = - \frac{1}{4} - \frac{7}{4} + 4 = - 2 + 4 = 2 \Rightarrow$ $F (x) = \frac{- 3}{x} + \frac{1}{x^{2}} + \frac{3}{x + 1} + \frac{2}{{(x + 1)}^{2}} + \frac{1}{{(x + 1)}^{3}}$ $S = \sum_{n = 1}^{\infty} {(- 1)}^{n} F (n) = - 3 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}$ $+ 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{3}} w e h a v e$ $\sum_{n = 1}^{\infty} \frac{x^{n}}{n} = - l n (1 - x) i f ∣ x ∣< 1 \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2)$ $i f δ (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{x}} w e h a v e p r o v e d t h a t δ (x) = (2^{1 - x} - 1) ξ (x)$ $\Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = (2^{- 1} - 1) ξ (2) = - \frac{1}{2} \frac{π^{2}}{6} = - \frac{π^{2}}{12}$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}$ $= - {- \frac{π^{2}}{12} + 1} = \frac{π^{2}}{12} - 1 a l s o$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{3}} = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{3}} = - \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{3}}$ $= - {δ (3) + 1} = - (2^{- 2} - 1) ξ (3) - 1 = - (- \frac{3}{4}) ξ (3) - 1$ $= \frac{3}{4} ξ (3) - 1 \Rightarrow$ $S = 3 l n (2) - \frac{π^{2}}{12} + \frac{2 π^{2}}{12} - 2 + \frac{3}{4} ξ (3) - 1$ $= 3 l n (2) + \frac{π^{2}}{12} + \frac{3}{4} ξ (3) - 3 .$
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