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Question Number 159233 by mnjuly1970 last updated on 14/Nov/21

Answered by mindispower last updated on 16/Nov/21

A_(2n) =Σ_(k≥1) (1/((2k)^(2n) ))=(1/2^(2n) ).ζ(2n) (((ζ(2n))/2^(2n) )/(2n(2n+1)))=−((ζ(2n))/(2^(2n) (2n+1)))+((ζ(2n))/(2^(2n) .(2n))) Σ_(k≥1) ζ(2k)x^(2k) =−((πx)/2)cot(πx)+(1/2) Σ_(k≥1) ((ζ(2k)x^(2k) )/(2k))=∫_0 ^x (−(π/2)cot(πx)+(1/(2x)))dx =lim_(t→0) ∫_t ^x (−(π/2)cot(πx)+(1/(2x)))dx= =lim_(t→0) ∫_t ^x (−(1/2)ln(sin(πx))+(1/2)ln(x)) (1/2)(ln((x/(sinπx)))−lim_(t→0) ln((t/(sin(πt)))) =ln((√(x/(sin(πx)))))+ln((√π)) Σ_(k≥1) ((ζ(2n))/(2^(2n) (2n)))=ln((√(1/2)))+ln((√π)) Σ_(n≥1) ((ζ(2n))/(2^(2n) (2n+1)))=2∫_0 ^(1/2) Σ_(n≥1) ζ(2n)x^(2n) dx =2∫_0 ^(1/2) ((1/2)−(π/2)xcot(πx))dx=(1/2)−(1/π)∫_0 ^(π/2) ucot(u)du =((1/2)+(1/π)∫_0 ^(π/2) ln(sin(u))du)=(1/2)+(1/π).−(π/2)ln(2)=(1/2)(1−ln(2)) S=(1/2)(ln(π)−ln(2))−(1/2)(1−ln(2)) =(1/2)(ln(π)−1)

$${A}_{\mathrm{2}{n}} =\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)^{\mathrm{2}{n}} }=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }.\zeta\left(\mathrm{2}{n}\right) \\ $$$$\frac{\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} }}{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}=−\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}+\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} .\left(\mathrm{2}{n}\right)} \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\sum}\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} =−\frac{\pi{x}}{\mathrm{2}}{cot}\left(\pi{x}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} }{\mathrm{2}{k}}=\int_{\mathrm{0}} ^{{x}} \left(−\frac{\pi}{\mathrm{2}}{cot}\left(\pi{x}\right)+\frac{\mathrm{1}}{\mathrm{2}{x}}\right){dx} \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{{t}} ^{{x}} \left(−\frac{\pi}{\mathrm{2}}{cot}\left(\pi{x}\right)+\frac{\mathrm{1}}{\mathrm{2}{x}}\right){dx}= \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{{t}} ^{{x}} \left(−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}\left(\pi{x}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}\right)\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\frac{{x}}{{sin}\pi{x}}\right)−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}{ln}\left(\frac{{t}}{{sin}\left(\pi{t}\right)}\right)\right. \\ $$$$={ln}\left(\sqrt{\frac{{x}}{{sin}\left(\pi{x}\right)}}\right)+{ln}\left(\sqrt{\pi}\right) \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}{n}\right)}={ln}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\right)+{ln}\left(\sqrt{\pi}\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \underset{{n}\geqslant\mathrm{1}} {\sum}\zeta\left(\mathrm{2}{n}\right){x}^{\mathrm{2}{n}} {dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}{xcot}\left(\pi{x}\right)\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ucot}\left({u}\right){du} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({u}\right)\right){du}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\pi}.−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{ln}\left(\mathrm{2}\right)\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\pi\right)−{ln}\left(\mathrm{2}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{ln}\left(\mathrm{2}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\pi\right)−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 15/Nov/21

very nice sir power...thx alot

$${very}\:{nice}\:{sir}\:{power}...{thx}\:{alot} \\ $$

Commented by mindispower last updated on 15/Nov/21

withe pleasur sir

$${withe}\:{pleasur}\:{sir} \\ $$

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