calculate Σ_(n=1) ^∞ (((2n+1)(−1)^n )/(n^2 (n+1)(n+2)^2 ))


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Question Number 69786 by mathmax by abdo last updated on 27/Sep/19

$c a l c u l a t e \sum_{n = 1}^{\infty} \frac{(2 n + 1) {(- 1)}^{n}}{n^{2} (n + 1) {(n + 2)}^{2}}$
Commented by mathmax by abdo last updated on 02/Oct/19
$let S_n =Σ_(k=1) ^n (((2k+1))/(k^2 (k+1)(k+2)^2 ))(−1)^k we have S=lim_(n→+∞) S_n let decompose F(x)=((2x+1)/(x^2 (x+1)(x+2)^2 )) F(x)=(a/x)+(b/x^2 ) +(c/(x+1)) +(d/(x+2)) +(e/((x+2)^2 )) b =lim_(x→0) x^2 F(x)=(1/4) e =lim_(x→−2) (x+2)^2 F(x) =((−3)/(−4))=(3/4) ⇒ F(x)=(a/x)+(1/(4x^2 )) +(c/(x+1)) +(d/(x+2)) +(3/(4(x+2)^2 )) lim_(x→+∞) xF(x)=0 =a+c+d F(1)=(3/(18)) =(1/6) =a+(1/4) +(c/2) +(d/3) +(1/(12)) =a+(c/2) +(d/3) +(1/3) ⇒ 1 =6a+3c+2d+2 ⇒6a+3c+2d=−1 F(−3) =((−5)/(9(−2))) =(5/(18)) =−(a/3)+(1/(36))−(c/2)−d+(3/4) =−(a/3)−(c/2)−d +(7/9) ⇒5=−6a−9c−18d+14 ⇒ −6a−9c−18d =−9 ⇒6a+9c+18d =9 ⇒2a+3c+6d =3 we get the system { ((a+c+d=0)),((6a+3c+2d =−1)) :} {2a+3c+6d =3 ⇒d=−a−c and { ((6a+3c+2(−a−c)=−1)),((2a+3c+6(−a−c)=3 ⇒ { ((4a+c=−1)),((−4a−3c =3 ⇒−2c=2 ⇒c=−1)) :})) :} 4a=−1+1 =0 ⇒a=0 ⇒d=1 ⇒ F(x)=(1/(4x^2 ))−(1/(x+1)) +(1/(x+2)) +(3/(4(x+2)^2 )) ⇒ S_n =Σ_(k=1) ^n (−1)^k F(k) =(1/4)Σ_(k=1) ^n (((−1)^k )/k^2 )−Σ_(k=1) ^n (((−1)^k )/(k+1)) +Σ_(k=1) ^n (((−1)^k )/(k+2)) +(3/4)Σ_(k=1) ^n (((−1)^k )/((k+2)^2 )) and lim_(n→+∞) S_n =(1/4)Σ_(k=1) ^∞ (((−1)^k )/k^2 )−Σ_(k=2) ^∞ (((−1)^(k−1) )/k) +Σ_(k=3) ^∞ (((−1)^(k−2) )/k)+(3/4)Σ_(k=3) ^∞ (((−1)^(k−2) )/k^2 ) Σ_(k=1) ^∞ (((−1)^k )/k^2 ) =(2^(1−2) −1)ξ(2) =−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(k=2) ^∞ (((−1)^(k−1) )/k) =−Σ_(k=2) ^∞ (((−1)^k )/k) =−(Σ_(k=1) ^∞ (((−1)^k )/k) +1) =−(−ln(2)+1) =ln(2)−1 Σ_(k=3) ^∞ (((−1)^(k−2) )/k) =Σ_(k=1) ^∞ (((−1)^k )/k) −(−1+(1/2))=−ln(2)+(1/2) Σ_(k=3) ^∞ (((−1)^(k−2) )/k^2 ) =Σ_(k=1) ^∞ (((−1)^k )/k^2 )−(−1+(1/2)) =−(π^2 /(12))+(1/2) ⇒ S=(1/4)(−(π^2 /(12)))−ln(2)+1−ln(2)+(1/2) +(3/4)(−(π^2 /(12))+(1/2)) =−(π^2 /(48))−2ln(2)+(3/2)+(3/8)−((3π^2 )/(48)) =((−4π^2 )/(48))−2ln(2)+((15)/8) =−(π^2 /(12))−2ln(2)+((15)/8)$
$l e t S_{n} = \sum_{k = 1}^{n} \frac{(2 k + 1)}{k^{2} (k + 1) {(k + 2)}^{2}} {(- 1)}^{k} w e h a v e S = {l i m}_{n \to + \infty} S_{n}$ $l e t d e c o m p o s e F (x) = \frac{2 x + 1}{x^{2} (x + 1) {(x + 2)}^{2}}$ $F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x + 1} + \frac{d}{x + 2} + \frac{e}{{(x + 2)}^{2}}$ $b = {l i m}_{x \to 0} x^{2} F (x) = \frac{1}{4}$ $e = {l i m}_{x \to - 2} {(x + 2)}^{2} F (x) = \frac{- 3}{- 4} = \frac{3}{4} \Rightarrow$ $F (x) = \frac{a}{x} + \frac{1}{4 x^{2}} + \frac{c}{x + 1} + \frac{d}{x + 2} + \frac{3}{4 {(x + 2)}^{2}}$ ${l i m}_{x \to + \infty} x F (x) = 0 = a + c + d$ $F (1) = \frac{3}{18} = \frac{1}{6} = a + \frac{1}{4} + \frac{c}{2} + \frac{d}{3} + \frac{1}{12} = a + \frac{c}{2} + \frac{d}{3} + \frac{1}{3} \Rightarrow$ $1 = 6 a + 3 c + 2 d + 2 \Rightarrow 6 a + 3 c + 2 d = - 1$ $F (- 3) = \frac{- 5}{9 (- 2)} = \frac{5}{18} = - \frac{a}{3} + \frac{1}{36} - \frac{c}{2} - d + \frac{3}{4}$ $= - \frac{a}{3} - \frac{c}{2} - d + \frac{7}{9} \Rightarrow 5 = - 6 a - 9 c - 18 d + 14 \Rightarrow$ $- 6 a - 9 c - 18 d = - 9 \Rightarrow 6 a + 9 c + 18 d = 9 \Rightarrow 2 a + 3 c + 6 d = 3$ $w e g e t t h e s y s t e m {\begin{cases} a + c + d = 0 \\ 6 a + 3 c + 2 d = - 1 \end{cases}$ ${2 a + 3 c + 6 d = 3 \Rightarrow d = - a - c a n d$ ${\begin{cases} 6 a + 3 c + 2 (- a - c) = - 1 \\ 2 a + 3 c + 6 (- a - c) = 3 \Rightarrow {\begin{cases} 4 a + c = - 1 \\ - 4 a - 3 c = 3 \Rightarrow - 2 c = 2 \Rightarrow c = - 1 \end{cases} \end{cases}$ $4 a = - 1 + 1 = 0 \Rightarrow a = 0 \Rightarrow d = 1 \Rightarrow$ $F (x) = \frac{1}{4 x^{2}} - \frac{1}{x + 1} + \frac{1}{x + 2} + \frac{3}{4 {(x + 2)}^{2}} \Rightarrow$ $S_{n} = \sum_{k = 1}^{n} {(- 1)}^{k} F (k) = \frac{1}{4} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2}} - \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k + 1} + \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k + 2}$ $+ \frac{3}{4} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{{(k + 2)}^{2}} a n d {l i m}_{n \to + \infty} S_{n}$ $= \frac{1}{4} \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k^{2}} - \sum_{k = 2}^{\infty} \frac{{(- 1)}^{k - 1}}{k} + \sum_{k = 3}^{\infty} \frac{{(- 1)}^{k - 2}}{k} + \frac{3}{4} \sum_{k = 3}^{\infty} \frac{{(- 1)}^{k - 2}}{k^{2}}$ $\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k^{2}} = (2^{1 - 2} - 1) ξ (2) = - \frac{1}{2} \frac{π^{2}}{6} = - \frac{π^{2}}{12}$ $\sum_{k = 2}^{\infty} \frac{{(- 1)}^{k - 1}}{k} = - \sum_{k = 2}^{\infty} \frac{{(- 1)}^{k}}{k} = - (\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} + 1)$ $= - (- l n (2) + 1) = l n (2) - 1$ $\sum_{k = 3}^{\infty} \frac{{(- 1)}^{k - 2}}{k} = \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} - (- 1 + \frac{1}{2}) = - l n (2) + \frac{1}{2}$ $\sum_{k = 3}^{\infty} \frac{{(- 1)}^{k - 2}}{k^{2}} = \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k^{2}} - (- 1 + \frac{1}{2})$ $= - \frac{π^{2}}{12} + \frac{1}{2} \Rightarrow$ $S = \frac{1}{4} (- \frac{π^{2}}{12}) - l n (2) + 1 - l n (2) + \frac{1}{2} + \frac{3}{4} (- \frac{π^{2}}{12} + \frac{1}{2})$ $= - \frac{π^{2}}{48} - 2 l n (2) + \frac{3}{2} + \frac{3}{8} - \frac{3 π^{2}}{48}$ $= \frac{- 4 π^{2}}{48} - 2 l n (2) + \frac{15}{8} = - \frac{π^{2}}{12} - 2 l n (2) + \frac{15}{8}$
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