Question Number 159233 by mnjuly1970 last updated on 14/Nov/21
Answered by mindispower last updated on 16/Nov/21
$${A}_{\mathrm{2}{n}} =\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)^{\mathrm{2}{n}} }=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }.\zeta\left(\mathrm{2}{n}\right) \\ $$$$\frac{\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} }}{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}=−\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}+\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} .\left(\mathrm{2}{n}\right)} \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\sum}\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} =−\frac{\pi{x}}{\mathrm{2}}{cot}\left(\pi{x}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} }{\mathrm{2}{k}}=\int_{\mathrm{0}} ^{{x}} \left(−\frac{\pi}{\mathrm{2}}{cot}\left(\pi{x}\right)+\frac{\mathrm{1}}{\mathrm{2}{x}}\right){dx} \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{{t}} ^{{x}} \left(−\frac{\pi}{\mathrm{2}}{cot}\left(\pi{x}\right)+\frac{\mathrm{1}}{\mathrm{2}{x}}\right){dx}= \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{{t}} ^{{x}} \left(−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}\left(\pi{x}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}\right)\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\frac{{x}}{{sin}\pi{x}}\right)−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}{ln}\left(\frac{{t}}{{sin}\left(\pi{t}\right)}\right)\right. \\ $$$$={ln}\left(\sqrt{\frac{{x}}{{sin}\left(\pi{x}\right)}}\right)+{ln}\left(\sqrt{\pi}\right) \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}{n}\right)}={ln}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\right)+{ln}\left(\sqrt{\pi}\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \underset{{n}\geqslant\mathrm{1}} {\sum}\zeta\left(\mathrm{2}{n}\right){x}^{\mathrm{2}{n}} {dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}{xcot}\left(\pi{x}\right)\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ucot}\left({u}\right){du} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({u}\right)\right){du}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\pi}.−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{ln}\left(\mathrm{2}\right)\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\pi\right)−{ln}\left(\mathrm{2}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{ln}\left(\mathrm{2}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\pi\right)−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 15/Nov/21
$${very}\:{nice}\:{sir}\:{power}…{thx}\:{alot} \\ $$
Commented by mindispower last updated on 15/Nov/21
$${withe}\:{pleasur}\:{sir} \\ $$