# 1-2-1-4-3-1-9-4-1-16-5-1-25-6-1-36-7-1-49-8-1-64-exp-2-2-pi-2-12-log-2-

Question Number 142035 by Dwaipayan Shikari last updated on 25/May/21
$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}/\mathrm{4}} }.\frac{\mathrm{3}^{\mathrm{1}/\mathrm{9}} }{\mathrm{4}^{\mathrm{1}/\mathrm{16}} }.\frac{\mathrm{5}^{\mathrm{1}/\mathrm{25}} }{\mathrm{6}^{\mathrm{1}/\mathrm{36}} }.\frac{\mathrm{7}^{\mathrm{1}/\mathrm{49}} }{\mathrm{8}^{\mathrm{1}/\mathrm{64}} }…={exp}\left(−\frac{\zeta'\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{log}\:\left(\mathrm{2}\right)\right) \\$$
Answered by mindispower last updated on 25/May/21
$$=\underset{{k}\geqslant\mathrm{1}} {\prod}\frac{\left(\mathrm{1}+\mathrm{2}{k}\right)^{\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }} }{\left(\mathrm{2}{k}\right)^{\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)^{\mathrm{2}} }} }=\Psi \\$$$${ln}\left(\Psi\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{ln}\left(\mathrm{2}{k}+\mathrm{1}\right)}{\left(\mathrm{2}{k}+\mathrm{1}\right)}−\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{ln}\left(\mathrm{2}{k}\right)}{\mathrm{4}{k}^{\mathrm{2}} }={A}−{B} \\$$$$\zeta\left({x}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{{x}} },\zeta'\left({x}\right)=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{ln}\left({n}\right)}{{n}^{{x}} } \\$$$${B}=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{4}}.\frac{\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{ln}\left({k}\right)}{{k}^{\mathrm{2}} }=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{4}}\zeta\left(\mathrm{2}\right)−\frac{\zeta'\left(\mathrm{2}\right)}{\mathrm{4}} \\$$$${A}=\underset{{k}\geqslant\mathrm{1}} {\sum}\left(\frac{{ln}\left({k}\right)}{{k}^{\mathrm{2}} }−\frac{{ln}\left(\mathrm{2}{k}\right)}{\mathrm{4}{k}^{\mathrm{2}} }\right)=−\zeta'\left(\mathrm{2}\right)−{B} \\$$$${ln}\left(\Psi\right)=−\frac{\mathrm{1}}{\mathrm{2}}\zeta'\left(\mathrm{2}\right)−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\zeta\left(\mathrm{2}\right)=−\frac{\zeta'\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{12}} \\$$$$\Psi={exp}\left(\frac{−\zeta'\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{12}}\right) \\$$$$\\$$
Commented by Dwaipayan Shikari last updated on 25/May/21
$${Thanks}\:{sir}.\:{Great}! \\$$
Commented by mindispower last updated on 26/May/21
$${pleasur}\:{sir} \\$$