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

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

$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{1}\:+\mathrm{2}^{\mathrm{3}} \:+\mathrm{3}^{\mathrm{3}} \:+...+{n}^{\mathrm{3}} } \\ $$

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)

$${we}\:{know}\:{that}\:\mathrm{1}\:+\mathrm{2}^{\mathrm{3}} \:+\mathrm{3}^{\mathrm{3}\:} \:+....+{n}^{\mathrm{3}} \:=\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{1}+\mathrm{2}^{\mathrm{3}} \:+\mathrm{3}^{\mathrm{3}} \:+...+{n}^{\mathrm{3}} }\:=\:\frac{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{find}\:{a}\:{and}\:{from}\:{R}\:/ \\ $$$$\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{{a}}{{n}^{\mathrm{2}} }\:\:+\frac{{b}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:={F}\left({n}\right) \\ $$$${a}\:={lim}_{{n}\rightarrow\mathrm{0}} {n}^{\mathrm{2}} {F}\left({n}\right)=\mathrm{1} \\ $$$${b}\:={lim}_{{n}\rightarrow−\mathrm{1}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({n}\right)=\:−\mathrm{1}\Rightarrow \\ $$$$\mathrm{4}\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\mathrm{4}\left\{\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right\}\:\:{we}\:{have} \\ $$$${S}\:={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:\:\:/ \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{1}+\mathrm{2}^{\mathrm{3}} \:+\mathrm{3}^{\mathrm{3}} \:+...+{k}^{\mathrm{3}} } \\ $$$${S}_{{n}} =\mathrm{4}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:−\mathrm{4}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$${S}_{{n}} =\mathrm{4}\:\xi_{{n}} \left(\mathrm{2}\right)\:−\mathrm{4}\:\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\xi_{{n}} \left(\mathrm{2}\right)\:−\mathrm{4}\left\{\:\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:−\mathrm{1}\right\} \\ $$$$\mathrm{4}\:\xi_{{n}} \left(\mathrm{2}\right)\:−\mathrm{4}\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:+\mathrm{4} \\ $$$${lim}_{{n}\rightarrow+\infty} \xi_{{n}} \left(\mathrm{2}\right)\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${lim}_{{n}\rightarrow+\infty} \xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow{lim}_{{n}\rightarrow\infty} {S}_{{n}} =\mathrm{4} \\ $$$${so}\:\:\:{S}\:=\mathrm{4} \\ $$$${let}\:{remind}\:{that}\:\xi_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{{x}} }\:.\left({x}>\mathrm{1}\right)\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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