Question Number 224288 by mnjuly1970 last updated on 31/Aug/25
$$ \\ $$$$\:\:\:\:\:\:\mathrm{P}=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{k}}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)}\:=?\:\:\:\: \\ $$$$ \\ $$
Commented by Frix last updated on 31/Aug/25
$$\mathrm{I}\:\mathrm{think}\:{P}=\sqrt{\pi} \\ $$
Commented by mr W last updated on 31/Aug/25
$${right}! \\ $$
Answered by mr W last updated on 31/Aug/25
![P_n =(((√1)×(√2)×...×(√n))/( (√2)×(√3)×...(√(n+1))))×((2×4×6×...×2n)/(1×3×5×...×(2n−1))) =(1/( (√(n+1))))×(((2×4×6×...×2n)^2 )/(1×2×3×4×5×6×...×(2n−1)×(2n))) =(1/( (√(n+1))))×(((2^n n!)^2 )/((2n)!)) ∼(1/( (√(n+1))))×((2^(2n) [(√(2πn))((n/e))^n ]^2 )/( (√(2π×2n))×(((2n)/e))^(2n) )) =((2^(2n) ×2πn)/( 2(√(πn(n+1)))×2^(2n) )) =(√(π/(1+(1/n)))) P=lim_(n→∞) P_n =lim_(n→∞) (√(π/(1+(1/n))))=(√π) ✓](https://www.tinkutara.com/question/Q224294.png)
$${P}_{{n}} =\frac{\sqrt{\mathrm{1}}×\sqrt{\mathrm{2}}×…×\sqrt{{n}}}{\:\sqrt{\mathrm{2}}×\sqrt{\mathrm{3}}×…\sqrt{{n}+\mathrm{1}}}×\frac{\mathrm{2}×\mathrm{4}×\mathrm{6}×…×\mathrm{2}{n}}{\mathrm{1}×\mathrm{3}×\mathrm{5}×…×\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$$\:\:=\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}}×\frac{\left(\mathrm{2}×\mathrm{4}×\mathrm{6}×…×\mathrm{2}{n}\right)^{\mathrm{2}} }{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}×\mathrm{6}×…×\left(\mathrm{2}{n}−\mathrm{1}\right)×\left(\mathrm{2}{n}\right)} \\ $$$$\:\:=\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}}×\frac{\left(\mathrm{2}^{{n}} {n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$\:\:\sim\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}}×\frac{\mathrm{2}^{\mathrm{2}{n}} \left[\sqrt{\mathrm{2}\pi{n}}\left(\frac{{n}}{{e}}\right)^{{n}} \right]^{\mathrm{2}} }{\:\sqrt{\mathrm{2}\pi×\mathrm{2}{n}}×\left(\frac{\mathrm{2}{n}}{{e}}\right)^{\mathrm{2}{n}} } \\ $$$$\:\:=\frac{\mathrm{2}^{\mathrm{2}{n}} ×\mathrm{2}\pi{n}}{\:\mathrm{2}\sqrt{\pi{n}\left({n}+\mathrm{1}\right)}×\mathrm{2}^{\mathrm{2}{n}} } \\ $$$$\:\:=\sqrt{\frac{\pi}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}} \\ $$$${P}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{P}_{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{\frac{\pi}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}}=\sqrt{\pi}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 01/Sep/25
$$\:\cancel{\lesseqgtr} \\ $$