Question Number 134657 by Dwaipayan Shikari last updated on 06/Mar/21
$$\frac{\mathrm{1}}{\pi{e}}+\frac{\mathrm{3}}{\mathrm{2}\pi^{\mathrm{2}} {e}^{\mathrm{2}} }+\frac{\mathrm{11}}{\mathrm{6}\pi^{\mathrm{3}} {e}^{\mathrm{3}} }+\frac{\mathrm{25}}{\mathrm{12}\pi^{\mathrm{4}} {e}^{\mathrm{4}} }+\frac{\mathrm{137}}{\mathrm{60}\pi^{\mathrm{5}} {e}^{\mathrm{5}} }+…=\frac{{log}\left(\pi\right)+\mathrm{1}−{log}\left(\pi{e}−\mathrm{1}\right)}{\pi{e}−\mathrm{1}} \\ $$