Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 148780 by Jonathanwaweh last updated on 31/Jul/21

Answered by Kamel last updated on 31/Jul/21

Ω=Π_(n=2) ^(+∞) e(1−(1/n^2 ))^n^2 =e^(Σ_(n=2) ^(+∞) (n^2 Ln(1−(1/n^2 ))+1)) =e^(Σ_(n=2) ^(+∞) n^2 (−Σ_(p=1) ^(+∞) (1/p)(1/n^(2p) )+(1/n^2 ))) =e^(−Σ_(p=1) ^(+∞) (1/(p+1))Σ_(n=2) ^(+∞) (1/n^(2p) )) =e^(−Σ_(p=1) ^(+∞) (1/(p+1))(ζ(2p)−1)) We have: Σ_(k=0) ^(+∞) ζ(2k)x^(2k) =−((πx)/2)cot(πx) So Σ_(k=0) ^(+∞) (ζ(2k)−1)x^(2k+1) =−((πx^2 )/2)cot(πx)−(x/(1−x^2 )) ∴ Σ_(p=0) ^(+∞) ((ζ(2p)−1)/(p+1))=−∫_0 ^1 (πx^2 cot(πx)+((2x)/(1−x^2 )))dx ∫(x^2 cot(πx)+((2x)/(1−x^2 )))dx=((ixLi_2 (e^(−2πix) ))/π^2 )+((Li_3 (e^(−2πix) ))/(2π^3 ))−((ix^3 )/π)+((x^2 Ln(1−e^(2πix) )−πLn(1−x^2 ))/π) ∴ ∫_0 ^1 (πx^2 cot(πx)+((2x)/(1−x^2 )))dx=Ln(π) So: Σ_(p=1) ^(+∞) ((ζ(2p)−1)/(p+1))=(3/2)−Ln(π) Then: 𝚷_(n=2) ^(+∞) e(1−(1/n^2 ))^n^2 =e^(Ln(𝛑)−(3/2)) =(𝛑/(e(√e))) KAMEL BENAICHA

$$ \\ $$$$\Omega=\underset{{n}=\mathrm{2}} {\overset{+\infty} {\prod}}{e}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{n}^{\mathrm{2}} } ={e}^{\underset{{n}=\mathrm{2}} {\overset{+\infty} {\sum}}\left({n}^{\mathrm{2}} {Ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\mathrm{1}\right)} \\ $$$$\:\:\:={e}^{\underset{{n}=\mathrm{2}} {\overset{+\infty} {\sum}}{n}^{\mathrm{2}} \left(−\underset{{p}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{p}}\frac{\mathrm{1}}{{n}^{\mathrm{2}{p}} }+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)} ={e}^{−\underset{{p}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{p}+\mathrm{1}}\underset{{n}=\mathrm{2}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}{p}} }} ={e}^{−\underset{{p}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{p}+\mathrm{1}}\left(\zeta\left(\mathrm{2}{p}\right)−\mathrm{1}\right)} \\ $$$${We}\:{have}:\:\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} =−\frac{\pi{x}}{\mathrm{2}}{cot}\left(\pi{x}\right) \\ $$$${So}\:\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\left(\zeta\left(\mathrm{2}{k}\right)−\mathrm{1}\right){x}^{\mathrm{2}{k}+\mathrm{1}} =−\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}{cot}\left(\pi{x}\right)−\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\therefore\:\:\underset{{p}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{p}\right)−\mathrm{1}}{{p}+\mathrm{1}}=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\pi{x}^{\mathrm{2}} {cot}\left(\pi{x}\right)+\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$$$\int\left({x}^{\mathrm{2}} {cot}\left(\pi{x}\right)+\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx}=\frac{{ixLi}_{\mathrm{2}} \left({e}^{−\mathrm{2}\pi{ix}} \right)}{\pi^{\mathrm{2}} }+\frac{{Li}_{\mathrm{3}} \left({e}^{−\mathrm{2}\pi{ix}} \right)}{\mathrm{2}\pi^{\mathrm{3}} }−\frac{{ix}^{\mathrm{3}} }{\pi}+\frac{{x}^{\mathrm{2}} {Ln}\left(\mathrm{1}−{e}^{\mathrm{2}\pi{ix}} \right)−\pi{Ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\pi} \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\pi{x}^{\mathrm{2}} {cot}\left(\pi{x}\right)+\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx}={Ln}\left(\pi\right) \\ $$$${So}:\:\underset{{p}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{p}\right)−\mathrm{1}}{{p}+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}}−{Ln}\left(\pi\right) \\ $$$${Then}:\:\:\:\:\:\:\:\:\:\:\:\underset{\boldsymbol{{n}}=\mathrm{2}} {\overset{+\infty} {\boldsymbol{\prod}}{e}}\left(\mathrm{1}−\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} }\right)^{\boldsymbol{{n}}^{\mathrm{2}} } =\boldsymbol{{e}}^{\boldsymbol{{Ln}}\left(\boldsymbol{\pi}\right)−\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\boldsymbol{\pi}}{\boldsymbol{{e}}\sqrt{\boldsymbol{{e}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{KAMEL}}\:\boldsymbol{{BENAICHA}} \\ $$

Commented by Tawa11 last updated on 02/Aug/21

Weldone sir

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

Answered by Kamel last updated on 31/Jul/21

Terms of Service

Privacy Policy

Contact: info@tinkutara.com