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Question Number 133988 by Dwaipayan Shikari last updated on 26/Feb/21

(2+(π/e))(((17)/(16))+(π/(4e)))(((82)/(81))+(π/(9e)))(((257)/(256))+(π/(16e)))...

$$\left(\mathrm{2}+\frac{\pi}{{e}}\right)\left(\frac{\mathrm{17}}{\mathrm{16}}+\frac{\pi}{\mathrm{4}{e}}\right)\left(\frac{\mathrm{82}}{\mathrm{81}}+\frac{\pi}{\mathrm{9}{e}}\right)\left(\frac{\mathrm{257}}{\mathrm{256}}+\frac{\pi}{\mathrm{16}{e}}\right)... \\ $$

Commented by Olaf last updated on 26/Feb/21

P = Π_(n=1) ^∞ (((?+1)/?)+(π/(n^2 e)))  ...I can′t find the sequence...  ...not enough data...

$$\mathrm{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{?+\mathrm{1}}{?}+\frac{\pi}{{n}^{\mathrm{2}} {e}}\right) \\ $$$$...{I}\:{can}'{t}\:{find}\:{the}\:{sequence}... \\ $$$$...{not}\:{enough}\:{data}... \\ $$

Commented by Dwaipayan Shikari last updated on 26/Feb/21

Π_(n=1) ^∞ (1+(1/n^4 )+(π/(en^2 ))).  Sorry i have edited

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{4}} }+\frac{\pi}{{en}^{\mathrm{2}} }\right).\:\:{Sorry}\:{i}\:{have}\:{edited}\: \\ $$

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