Question-173721

Question Number 173721 by KONE last updated on 16/Jul/22

Answered by mnjuly1970 last updated on 17/Jul/22

e^( ln(lim_(n→∞) (Π_(k=1) ^n (e)^(1/n) ((2/π) )^(1/k^( 2) ) ) )) = e^( lim_( n→∞) (Σ_(k=1) ^n ln((e)^(1/n) ((2/π) )^( (1/k^( 2) )) ))) = e^( lim_( n→∞) (Σ_(k=1) ^n (1/n) + ln((2/π))Σ_(k=1) ^n (1/k^( 2) ))) = e^( ( 1 +(π^( 2) /6) ln((2/π)))= e . ((2/π) )^( (π^( 2) /6)) )

$$\:\:{e}^{\:\mathrm{ln}\left(\mathrm{lim}_{{n}\rightarrow\infty} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\sqrt[{{n}}]{{e}}\:\left(\frac{\mathrm{2}}{\pi}\:\right)^{\frac{\mathrm{1}}{{k}^{\:\mathrm{2}} }} \right)\:\right)} \\ $$$$\:\:\:\:\:\:=\:{e}^{\:\mathrm{lim}_{\:{n}\rightarrow\infty} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{ln}\left(\sqrt[{{n}}]{{e}}\:\:\left(\frac{\mathrm{2}}{\pi}\:\right)^{\:\frac{\mathrm{1}}{{k}^{\:\mathrm{2}} }} \right)\right)} \\ $$$$\:\:\:\:\:\:\:=\:{e}^{\:\mathrm{lim}_{\:{n}\rightarrow\infty} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}}\:+\:\mathrm{ln}\left(\frac{\mathrm{2}}{\pi}\right)\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\:\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\:\:\:=\:{e}^{\:\left(\:\mathrm{1}\:+\frac{\pi^{\:\mathrm{2}} }{\mathrm{6}}\:\mathrm{ln}\left(\frac{\mathrm{2}}{\pi}\right)\right)=\:{e}\:\:.\:\left(\frac{\mathrm{2}}{\pi}\:\right)^{\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{6}}} \:} \\ $$

Commented by KONE last updated on 17/Jul/22

$${merci}\:{bien}\:{a}\:{vous} \\ $$

Commented by mnjuly1970 last updated on 18/Jul/22

$${you}\:{are}\:{welcom} \\ $$

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