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e-pi-1-e-pi-1-pi-2-pi-2-6-pi-2-10-pi-2-14-




Question Number 126704 by Dwaipayan Shikari last updated on 23/Dec/20
((e^π −1)/(e^π +1))=(π/(2+(π^2 /(6+(π^2 /(10+(π^2 /(14+....))))))))
$$\frac{{e}^{\pi} −\mathrm{1}}{{e}^{\pi} +\mathrm{1}}=\frac{\pi}{\mathrm{2}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}+\frac{\pi^{\mathrm{2}} }{\mathrm{10}+\frac{\pi^{\mathrm{2}} }{\mathrm{14}+….}}}} \\ $$

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