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


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Question Number 37339 by math khazana by abdo last updated on 12/Jun/18

$f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{2 n + 1}{1 + 2^{3} + 3^{3} + . . . + n^{3}}$
Commented by math khazana by abdo last updated on 13/Jun/18
$we know that 1 +2^3 +3^(3 ) +....+n^3 =((n^2 (n+1)^2 )/4) ⇒ ((2n+1)/(1+2^3 +3^3 +...+n^3 )) = ((4(2n+1))/(n^2 (n+1)^2 )) let find a and from R / ((2n+1)/(n^2 (n+1)^2 )) = (a/n^2 ) +(b/((n+1)^2 )) =F(n) a =lim_(n→0) n^2 F(n)=1 b =lim_(n→−1) (n+1)^2 F(n)= −1⇒ 4 ((2n+1)/(n^2 (n+1)^2 )) =4{ (1/n^2 ) −(1/((n+1)^2 ))} we have S =lim_(n→+∞) S_n / S_n =Σ_(k=1) ^n ((2k+1)/(1+2^3 +3^3 +...+k^3 )) S_n =4 Σ_(k=1) ^n (1/k^2 ) −4 Σ_(k=1) ^n (1/((k+1)^2 )) but S_n =4 ξ_n (2) −4 Σ_(k=2) ^(n+1) (1/k^2 ) =4ξ_n (2) −4{ ξ_(n+1) (2) −1} 4 ξ_n (2) −4ξ_(n+1) (2) +4 lim_(n→+∞) ξ_n (2) =Σ_(k=1) ^∞ (1/k^2 ) =(π^2 /6) lim_(n→+∞) ξ_(n+1) (2) =Σ_(k=1) ^∞ (1/k^2 ) =(π^2 /6) ⇒lim_(n→∞) S_n =4 so S =4 let remind that ξ_n (x)=Σ_(k=1) ^n (1/k^x ) .(x>1)$
$w e k n o w t h a t 1 + 2^{3} + 3^{3} + . . . . + n^{3} = \frac{n^{2} {(n + 1)}^{2}}{4}$ $\Rightarrow \frac{2 n + 1}{1 + 2^{3} + 3^{3} + . . . + n^{3}} = \frac{4 (2 n + 1)}{n^{2} {(n + 1)}^{2}}$ $l e t f i n d a a n d f r o m R /$ $\frac{2 n + 1}{n^{2} {(n + 1)}^{2}} = \frac{a}{n^{2}} + \frac{b}{{(n + 1)}^{2}} = F (n)$ $a = {l i m}_{n \to 0} n^{2} F (n) = 1$ $b = {l i m}_{n \to - 1} {(n + 1)}^{2} F (n) = - 1 \Rightarrow$ $4 \frac{2 n + 1}{n^{2} {(n + 1)}^{2}} = 4 {\frac{1}{n^{2}} - \frac{1}{{(n + 1)}^{2}}} w e h a v e$ $S = {l i m}_{n \to + \infty} S_{n} /$ $S_{n} = \sum_{k = 1}^{n} \frac{2 k + 1}{1 + 2^{3} + 3^{3} + . . . + k^{3}}$ $S_{n} = 4 \sum_{k = 1}^{n} \frac{1}{k^{2}} - 4 \sum_{k = 1}^{n} \frac{1}{{(k + 1)}^{2}} b u t$ $S_{n} = 4 ξ_{n} (2) - 4 \sum_{k = 2}^{n + 1} \frac{1}{k^{2}}$ $= 4 ξ_{n} (2) - 4 {ξ_{n + 1} (2) - 1}$ $4 ξ_{n} (2) - 4 ξ_{n + 1} (2) + 4$ ${l i m}_{n \to + \infty} ξ_{n} (2) = \sum_{k = 1}^{\infty} \frac{1}{k^{2}} = \frac{π^{2}}{6}$ ${l i m}_{n \to + \infty} ξ_{n + 1} (2) = \sum_{k = 1}^{\infty} \frac{1}{k^{2}} = \frac{π^{2}}{6} \Rightarrow {l i m}_{n \to \infty} S_{n} = 4$ $s o S = 4$ $l e t r e m i n d t h a t ξ_{n} (x) = \sum_{k = 1}^{n} \frac{1}{k^{x}} . (x > 1)$
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