Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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

$${pls}\:{how}\:{can}\:{i}\:{contact}\:{you}\:{on}\:{whatsapp}\:{sir}\:\:{mind}\:{is}\:{power} \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com