find the value of Σ_(n=1) ^∞ (((−1)^n )/(n^3 (n+1)^4 ))


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Question Number 60499 by abdo mathsup 649 cc last updated on 21/May/19

$f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{3} {(n + 1)}^{4}}$
Commented by maxmathsup by imad last updated on 30/May/19

$l e t d e c o m p o s e F (x) = \frac{1}{x^{3} {(x + 1)}^{4}} w e h a v e$ $F (x) = \sum_{i = 1}^{3} \frac{a_{i}}{x^{i}} + \sum_{i = 1}^{4} \frac{b_{i}}{{(x + 1)}^{i}} l e t d e t e r m i n e D_{2} (0) f o r$ $f (x) = {(x + 1)}^{- 4} w e h a v e f (x) = f (0) + \frac{f^{^{'}} (0)}{1!} x + \frac{f^{(2)} (0)}{2!} x^{2} + \frac{x^{3}}{3!} ξ (x)$ $f (0) = 1, f^{^{'}} (x) = - 4 {(x + 1)}^{- 5} \Rightarrow f^{^{'}} (0) = - 4, f^{(2)} (x) = 20 {(x + 1)}^{- 6} \Rightarrow$ $f^{(2)} (0) = 20 \Rightarrow f (x) = 1 - 4 x + 10 x^{2} + x^{3} ξ (x) \Rightarrow$ $\frac{f (x)}{x^{3}} = \frac{1}{x^{3}} - \frac{4}{x^{2}} + \frac{10}{x} + ξ (x) \Rightarrow a_{1} = 10, a_{2} = - 4, a_{3} = - 1 a l s o w e h a v e$ $F (x) =_{x + 1 = t} \frac{1}{{(t - 1)}^{3} t^{4}} l e t f i n d D_{3} (0) f o r g (t) = {(t - 1)}^{- 3}$ $g (t) = g (0) + \frac{g^{^{'}} (0)}{1!} t + \frac{g^{(2)} (0)}{2!} t^{2} + \frac{g^{(3)} (0)}{3!} t^{3} + \frac{t^{4}}{4!} ξ (t)$ $g (0) = - 1, g^{^{'}} (t) = - 3 {(t - 1)}^{- 4} \Rightarrow g^{^{'}} (0) = - 3$ $g^{(2)} (t) = 12 {(t - 1)}^{- 5} \Rightarrow g^{(2)} (0) = - 12, g^{(3)} (t) = - 60 {(t - 1)}^{- 6} \Rightarrow$ $g^{(3)} (0) = - 60 \Rightarrow$ $g (t) = - 1 - 3 t - 6 t^{2} - 10 t^{3} + \frac{t^{4}}{4!} ξ (t) \Rightarrow$ $\frac{g (t)}{t^{4}} = - \frac{1}{t^{4}} - \frac{3}{t^{3}} - \frac{6}{t^{2}} - \frac{10}{t} + \frac{1}{4!} ξ (t)$ $= - \frac{1}{{(x + 1)}^{4}} - \frac{3}{{(x + 1)}^{3}} - \frac{6}{{(x + 1)}^{2}} - \frac{10}{(x + 1)} + \frac{1}{4!} ξ (x + 1) \Rightarrow$ $b_{1} = - 10, b_{2} = - 6, b_{3} = - 3, b_{4} = - 1 a n d$ $F (x) = \frac{10}{x} - \frac{4}{x^{2}} - \frac{1}{x^{3}} - \frac{10}{(x + 1)} - \frac{6}{{(x + 1)}^{2}} - \frac{3}{{(x + 1)}^{3}} - \frac{1}{{(x + 1)}^{4}} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{3} {(n + 1)}^{4}} = \sum_{n = 1}^{\infty} {(- 1)}^{n} F (n)$ $= 10 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} - 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} - 10 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} - 6 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}}$ $- 3 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{3}} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{4}}$
Commented by maxmathsup by imad last updated on 30/May/19
$Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4)ξ(2) =(3/4) (π^2 /6) =(π^2 /8) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(24)) −(π^2 /8) =((−2π^2 )/(24)) =−(π^2 /(12)) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n) =−Σ_(n=2) ^∞ (((−1)^n )/n) =−{Σ_(n=1) ^∞ (((−1)^n )/n) +1) =−(−ln(2))−1 =ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^2 ) =−{ Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +1) =−(−(π^2 /(12)))−1 =(π^2 /(12)) −1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^3 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^3 ) for that let calculate δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) interms of ξ(x) =Σ_(n=1) ^∞ (1/n^x ) (x>1) we have δ(x) =Σ_(n=1) ^∞ (1/(2^x n^x )) − Σ_(n=0) ^∞ (1/((2n+1)^x )) ξ(x) =Σ_(n=1) ^∞ (1/n^x ) =(1/2^x ) ξ(x) +Σ_(n=0) ^∞ (1/((2n+1)^x )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^x )) =(1−2^(−x) )ξ(x) ⇒ δ(x) =2^(−x) ξ(x)−(1−2^(−x) )ξ(x) =(2^(1−x) −1)ξ(x) Σ_(n=1) ^∞ (((−1)^n )/n^3 ) =(2^(−2) −1)ξ(3)=((1/4)−1)ξ(3) =−(3/4)ξ(3) Σ_(n=1) ^∞ (((−1)^n )/n^4 ) =(2^(−3) −1)ξ(4) =((1/8) −1)ξ(4) =−(7/8)ξ(4) so the value of S is determined...$
$\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2)$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}$ $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{3}{4} ξ (2)$ $= \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \frac{π^{2}}{24} - \frac{π^{2}}{8} = \frac{- 2 π^{2}}{24} = - \frac{π^{2}}{12}$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = - \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n}$ $= - {\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} + 1) = - (- l n (2)) - 1 = l n (2) - 1$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}$ $= - {\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} + 1) = - (- \frac{π^{2}}{12}) - 1 = \frac{π^{2}}{12} - 1$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{3}} = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{3}} = - \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{3}}$ $f o r t h a t l e t c a l c u l a t e δ (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{x}} i n t e r m s o f$ $ξ (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{x}} (x > 1) w e h a v e$ $δ (x) = \sum_{n = 1}^{\infty} \frac{1}{2^{x} n^{x}} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{x}}$ $ξ (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{x}} = \frac{1}{2^{x}} ξ (x) + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{x}} \Rightarrow$ $\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{x}} = (1 - 2^{- x}) ξ (x) \Rightarrow$ $δ (x) = 2^{- x} ξ (x) - (1 - 2^{- x}) ξ (x) = (2^{1 - x} - 1) ξ (x)$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{3}} = (2^{- 2} - 1) ξ (3) = (\frac{1}{4} - 1) ξ (3) = - \frac{3}{4} ξ (3)$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{4}} = (2^{- 3} - 1) ξ (4) = (\frac{1}{8} - 1) ξ (4) = - \frac{7}{8} ξ (4)$ $s o t h e v a l u e o f S i s d e t e r m i n e d . . .$
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