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


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Question Number 41408 by maxmathsup by imad last updated on 06/Aug/18

$c a l c u l a t e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n (n + 1) (n + 2) (n + 3)}$
Commented by maxmathsup by imad last updated on 07/Aug/18
$let decompose F(x)= (1/(x(x+1)(x+2)(x+3))) F(x)=(a/x) +(b/(x+1)) +(c/(x+2)) +(d/(x+3)) a =lim_(x→0) xF(x)=(1/6) b =lim_(x→−1) (x+1)F(x) = −(1/2) c =lim_(x→−2) (x+2)F(x) = (1/2) d =lim_(x→−3) (x+3)F(x)=−(1/6) ⇒ F(x)=(1/(6x)) −(1/(2(x+1))) +(1/(2(x+2))) −(1/(6(x+3))) ⇒ S =(1/6) Σ_(n=1) ^∞ (((−1)^n )/n) −(1/2)Σ_(n=1) ^∞ (((−1)^n )/(n+1)) +(1/2) Σ_(n=1) ^∞ (((−1)^n )/(n+2)) −(1/6)Σ_(n=1) ^∞ (((−1)^n )/(n+3)) but Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n) =ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/(n+2)) =Σ_(n=3) ^∞ (((−1)^(n−2) )/n) =Σ_(n=3) ^∞ (((−1)^n )/n) =−ln(2)−{−1+(1/2)} =−ln(2)+(1/2) Σ_(n=1) ^∞ (((−1)^n )/(n+3)) =Σ_(n=4) ^∞ (((−1)^(n−3) )/n) =Σ_(n=4) ^∞ (((−1)^(n−1) )/n) =ln(2)−{1−(1/2) +(1/3)} =ln(2)−((1/2) +(1/3))=ln(2)−(5/6) ⇒ S =−(1/6)ln(2) −(1/2)ln(2) +(1/2) −(1/2)ln(2) +(1/4) −(1/6)ln(2) +(5/(36)) S =(−(1/3) −1)ln(2) +(3/4) +(5/(36)) =−(4/3)ln(2) + ((27+5)/(36)) =−(4/3)ln(2) + ((16)/(18)) S =(8/9) −(4/3)ln(2)$
$l e t d e c o m p o s e F (x) = \frac{1}{x (x + 1) (x + 2) (x + 3)}$ $F (x) = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{x + 2} + \frac{d}{x + 3}$ $a = {l i m}_{x \to 0} x F (x) = \frac{1}{6}$ $b = {l i m}_{x \to - 1} (x + 1) F (x) = - \frac{1}{2}$ $c = {l i m}_{x \to - 2} (x + 2) F (x) = \frac{1}{2}$ $d = {l i m}_{x \to - 3} (x + 3) F (x) = - \frac{1}{6} \Rightarrow$ $F (x) = \frac{1}{6 x} - \frac{1}{2 (x + 1)} + \frac{1}{2 (x + 2)} - \frac{1}{6 (x + 3)} \Rightarrow$ $S = \frac{1}{6} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} - \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} + \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 2} - \frac{1}{6} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 3} b u t$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2)$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2) - 1$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 2} = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n - 2}}{n} = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2) - {- 1 + \frac{1}{2}}$ $= - l n (2) + \frac{1}{2}$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 3} = \sum_{n = 4}^{\infty} \frac{{(- 1)}^{n - 3}}{n} = \sum_{n = 4}^{\infty} \frac{{(- 1)}^{n - 1}}{n}$ $= l n (2) - {1 - \frac{1}{2} + \frac{1}{3}} = l n (2) - (\frac{1}{2} + \frac{1}{3}) = l n (2) - \frac{5}{6} \Rightarrow$ $S = - \frac{1}{6} l n (2) - \frac{1}{2} l n (2) + \frac{1}{2} - \frac{1}{2} l n (2) + \frac{1}{4} - \frac{1}{6} l n (2) + \frac{5}{36}$ $S = (- \frac{1}{3} - 1) l n (2) + \frac{3}{4} + \frac{5}{36} = - \frac{4}{3} l n (2) + \frac{27 + 5}{36} = - \frac{4}{3} l n (2) + \frac{16}{18}$ $S = \frac{8}{9} - \frac{4}{3} l n (2)$
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