Question Number 141694 by Dwaipayan Shikari last updated on 22/May/21
$${log}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{10}}\mathrm{9}{e}^{\gamma} \right)=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }\right)−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{3}}\:\left(\frac{\mathrm{1}^{\mathrm{3}} +\mathrm{9}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }\:\right)+\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{4}}\left(\frac{\mathrm{1}^{\mathrm{4}} +\mathrm{9}^{\mathrm{4}} }{\mathrm{10}^{\mathrm{4}} }\right)−… \\ $$$$\gamma={Euler}\:{Mascheroni}\:{Constant} \\ $$