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Question-75403

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Question Number 75403 by TawaTawa last updated on 10/Dec/19
Answered by mind is power last updated on 10/Dec/19
=(Σ_(m≥0) (1/((2m+1)^6 )))(Σ_(n=1) ^(+∞) (((−1)^(n+1) )/(4n^2 −1))) Σ_(n≥1) (((−1)^(n+1) )/((2n−1)(2n+1)))=Σ_(n≥1) (((−1)^(n+1) )/(4n^2 −1)) =(1/2){Σ_(n≥1) (((−1)^(n+1) )/((2n−1)))+Σ_(n≥1) (((−1)^(n+2) )/(2n+1))} =(1/2)+{Σ_(n≥1) (((−1)^(n+2) )/(2n+1))} =(1/2)+Σ_(n≥1) (((−1)^n )/(2n+1)) (1/(1+x^2 ))=Σ_(k≥0) (−x^2 )^k ⇒∫(1/(1+x^2 ))=Σ_(k≥0) (((−1)^k )/(2k+1))x^(2k+1) ⇒Σ_(k≥0) (((−1)^k )/(2k+1))=arctan(1)=(π/4) ⇒Σ_(k≥1) (((−1)^k )/(2k+1))=(π/4)−1 ⇒Σ_(n≥1) (1/(4n^2 −1))=(π/4)−(1/2)=((π−2)/4) ζ(6)=Σ_(k≥1) (1/k^6 )=(π^6 /(945)) ⇒Σ_(k≥1) (1/((2k+1)^6 ))=ζ(6)−(1/2^6 )ζ(6)=(((2^6 −1)/2^6 )).(π^6 /(945))=((63π^6 )/(64.945))=(π^6 /(960)) ⇒Σ_(m≥0) Σ_(n≥1) (1/((2m+1)^6 )).(((−1)^(n+1) )/(4n^2 −1))=(π^6 /(960)).((π−2)/4)=((π^6 (π−2))/(3840))
$$=\left(\underset{\mathrm{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2m}+\mathrm{1}\right)^{\mathrm{6}} }\right)\left(\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{4n}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\left(\mathrm{2n}−\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{4n}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\left(\mathrm{2n}−\mathrm{1}\right)}+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{2}} }{\mathrm{2n}+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\left\{\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{2}} }{\mathrm{2n}+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{k}} \Rightarrow\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2k}+\mathrm{1}}\mathrm{x}^{\mathrm{2k}+\mathrm{1}} \\ $$$$\Rightarrow\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2k}+\mathrm{1}}=\mathrm{arctan}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2k}+\mathrm{1}}=\frac{\pi}{\mathrm{4}}−\mathrm{1} \\ $$$$\Rightarrow\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{4n}^{\mathrm{2}} −\mathrm{1}}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi−\mathrm{2}}{\mathrm{4}} \\ $$$$\zeta\left(\mathrm{6}\right)=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{6}} }=\frac{\pi^{\mathrm{6}} }{\mathrm{945}} \\ $$$$\Rightarrow\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{6}} }=\zeta\left(\mathrm{6}\right)−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\zeta\left(\mathrm{6}\right)=\left(\frac{\mathrm{2}^{\mathrm{6}} −\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\right).\frac{\pi^{\mathrm{6}} }{\mathrm{945}}=\frac{\mathrm{63}\pi^{\mathrm{6}} }{\mathrm{64}.\mathrm{945}}=\frac{\pi^{\mathrm{6}} }{\mathrm{960}} \\ $$$$\Rightarrow\underset{\mathrm{m}\geqslant\mathrm{0}} {\sum}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2m}+\mathrm{1}\right)^{\mathrm{6}} }.\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{4n}^{\mathrm{2}} −\mathrm{1}}=\frac{\pi^{\mathrm{6}} }{\mathrm{960}}.\frac{\pi−\mathrm{2}}{\mathrm{4}}=\frac{\pi^{\mathrm{6}} \left(\pi−\mathrm{2}\right)}{\mathrm{3840}} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 10/Dec/19
God bless you sir, i appreciate your time
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$
Commented by mind is power last updated on 10/Dec/19
y′re Welcom
$$\mathrm{y}'\mathrm{re}\:\mathrm{Welcom} \\ $$
Commented by feli last updated on 11/Dec/19
pls how can i contact you on whatsapp sir mind is power
$${pls}\:{how}\:{can}\:{i}\:{contact}\:{you}\:{on}\:{whatsapp}\:{sir}\:\:{mind}\:{is}\:{power} \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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