Question Number 196429 by SANOGO last updated on 24/Aug/23
Answered by Frix last updated on 24/Aug/23
$$\mathrm{ln}\:\left(\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\:\left(\sqrt[{{n}}]{\mathrm{e}}\left(\frac{\mathrm{2}}{\pi}\right)^{\frac{\mathrm{1}}{{k}^{\mathrm{2}} }} \right)\right)\:=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{ln}\:\mathrm{2}\:−\mathrm{ln}\:\pi}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}}\right)\:= \\ $$$$=\mathrm{1}+\mathrm{ln}\:\frac{\mathrm{2}}{\pi}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\mathrm{ln}\:\frac{\mathrm{2}}{\pi} \\ $$$$\mathrm{e}^{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\mathrm{ln}\:\frac{\mathrm{2}}{\pi}} =\left(\frac{\mathrm{2}}{\pi}\right)^{\frac{\pi^{\mathrm{2}} }{\mathrm{6}}} \mathrm{e} \\ $$
Commented by SANOGO last updated on 24/Aug/23
$${thank}\:{you} \\ $$
Commented by Frix last updated on 24/Aug/23
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