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Question Number 205142 by universe last updated on 10/Mar/24

$$\underset{n\to \mathrm{\infty }}{\lim }{\mathrm{n}}^{-{\mathrm{n}}^2}{[(\mathrm{n}+1)(\mathrm{n}+\frac{1}{2})(\mathrm{n}+\frac{1}{2^2})...(\mathrm{n}+\frac{1}{2^{\mathrm{n}-1}})]}^{\mathrm{n}}=?$$

Answered by pi314 last updated on 10/Mar/24

$$e^{-n^{2}ln\left(n\right))n\underset{k=0}{\sum ^{n-1}}ln(n+\frac{1}{2^k})}=e^{-n^{2}ln\left(n\right)+n\underset{k=0}{\sum ^{n-1}}ln\left(n\right)+ln(1+\frac{1}{2^{k}n})}$$$$=e^{ln(1+\frac{1}{n{2}^k})}$$$$0⩽ln(1+x)⩽x;\mathrm{\forall }x⩾0$$$$ln(1+\frac{1}{n{2}^k})⩽\frac{1}{n{2}^k}$$$$\Rightarrow \underset{k=0}{\sum ^{n-1}}\frac{1}{n.2^k}=\frac{1}{n}\left(\frac{1-{\left(\frac{1}{2}\right)}^n}{\frac{1}{2}}\right)=\frac{2^n-1}{n.2^{n-1}}\to 0$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\underset{k=0}{\sum ^{n-1}}ln(1+\frac{1}{n{2}^k})=0$$$$e^{ln(n^{-n^2}{[(n+1)....(n+\frac{1}{2^{n-1}})]}^n}=e^{\underset{k=0}{\sum ^{n-1}}ln(1+\frac{1}{n{2}^k})}\stackrel{\mathrm{\infty }}{\to }{e}^0=1$$

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