Question Number 205142 by universe last updated on 10/Mar/24
![lim_(n→∞) n^(−n^2 ) [(n+1)(n+(1/2))(n+(1/2^2 ))...(n+(1/2^(n−1) ))]^n =?](https://www.tinkutara.com/question/Q205142.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}^{−\mathrm{n}^{\mathrm{2}} } \left[\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)…\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\right)\right]^{\mathrm{n}} =? \\ $$
Answered by pi314 last updated on 10/Mar/24
![e^(−n^2 ln(n))nΣ_(k=0) ^(n−1) ln(n+(1/2^k ))) =e^(−n^2 ln(n)+nΣ_(k=0) ^(n−1) ln(n)+ln(1+(1/(2^k n)))) =e^(ln(1+(1/(n2^k )))) 0≤ln(1+x)≤x;∀x≥0 ln(1+(1/(n2^k )))≤(1/(n2^k )) ⇒Σ_(k=0) ^(n−1) (1/(n.2^k ))=(1/n)(((1−((1/2))^n )/(1/2)))=((2^n −1)/(n.2^(n−1) ))→0 ⇒lim_(n→∞) Σ_(k=0) ^(n−1) ln(1+(1/(n2^k )))=0 e^(ln(n^(−n^2 ) [(n+1)....(n+(1/2^(n−1) ))]^n ) =e^(Σ_(k=0) ^(n−1) ln(1+(1/(n2^k )))) →^∞ e^0 =1](https://www.tinkutara.com/question/Q205143.png)
$${e}^{\left.−{n}^{\mathrm{2}} {ln}\left({n}\right)\right){n}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left({n}+\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\right)} ={e}^{−{n}^{\mathrm{2}} {ln}\left({n}\right)+{n}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left({n}\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{{k}} {n}}\right)} \\ $$$$={e}^{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}\mathrm{2}^{{k}} }\right)} \\ $$$$\mathrm{0}\leqslant{ln}\left(\mathrm{1}+{x}\right)\leqslant{x};\forall{x}\geqslant\mathrm{0} \\ $$$${ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}\mathrm{2}^{{k}} }\right)\leqslant\frac{\mathrm{1}}{{n}\mathrm{2}^{{k}} } \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{n}.\mathrm{2}^{{k}} }=\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} }{\frac{\mathrm{1}}{\mathrm{2}}}\right)=\frac{\mathrm{2}^{{n}} −\mathrm{1}}{{n}.\mathrm{2}^{{n}−\mathrm{1}} }\rightarrow\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}\mathrm{2}^{{k}} }\right)=\mathrm{0} \\ $$$${e}^{{ln}\left({n}^{−{n}^{\mathrm{2}} } \left[\left({n}+\mathrm{1}\right)….\left({n}+\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)\right]^{{n}} \right.} ={e}^{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}\mathrm{2}^{{k}} }\right)} \overset{\infty} {\rightarrow}{e}^{\mathrm{0}} =\mathrm{1} \\ $$