$\lim_{n\rightarrow \infty}\left[\frac{\left(1+\frac{1}{n^2}\right)\cdot \cdots\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n$

Question

$$\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$

Try: $$y=\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$

$$\log_{e}(y) =n\sum^{n}_{r=1}\log_{e}\bigg(1+\frac{r}{n^2}\bigg)-\frac{n}{2}$$

could some help me to solve it , thanks

as @RonGordon said, use the power series expansion of $\ln(1+x)$ for $|x|<1$ to sandwich the limit — Masacroso, Jan 04 '18 at 14:23
@Ron Gordon Sorry, you're wrong: the second term of the Taylor expansion is not small enough. — , Jan 04 '18 at 14:44
@ProfessorVector: OK, Please explain why, for $n$ sufficiently large, $\log{\left ( 1+\frac{1}{n} \right )}$ is not close in value to $\frac{1}{n}$. — Ron Gordon, Jan 04 '18 at 14:47
@Ron Gordon We aren't speaking of $\log{\left ( 1+\frac{1}{n} \right )}$. — , Jan 04 '18 at 14:52

score 6 · Accepted Answer · answered Jan 04 '18 at 15:01

We have $$\log{\left ( 1+\frac{r}{n^2} \right )}=\frac{r}{n^2}-\frac12\frac{r^2}{n^4}+O\left(\frac{r^3}{n^6}\right),$$ that means $$n\sum^n_ {r=1}\log{\left ( 1+\frac{r}{n^2} \right )}=\frac{n(n+1)}2\frac1n-\frac12\frac{n(n+1)(2n+1)}6\frac1{n^3}+O(n^4)\frac1{n^5},$$ so we get $$n\sum^n_ {r=1}\log{\left ( 1+\frac{r}{n^2} \right )}-\frac{n}2\to\frac12-\frac16=\frac13$$ as $n\to\infty$.

Robert Z · Answer 2 · 2018-01-04T15:47:58.120

We have that $$\begin{align} \left[\frac{\left(1+\frac{1}{n^2}\right)\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n&=\frac{\left(\frac{(n^2+n)!}{(n^2)!}\right)^n}{n^{2n^2} e^{n/2}} \sim\frac{\left(1+\frac{1}{n}\right)^{n^3+n^2+\frac{n}{2}}}{e^{n^2+n/2}} \\ &\sim\frac{\exp\left((n^3+n^2+\frac{n}{2})\ln\left(1+\frac{1}{n}\right)\right)}{e^{n^2+n/2}}\\ &\sim\frac{\exp\left((n^2+n+\frac{1}{2})-\frac{n+1}{2}+\frac{1}{3}\right)}{e^{n^2+n/2}}\to e^{1/3} \end{align}$$ where we used the Stirling approximation for factorials and the expansion $$\ln(1+t)= t-\frac{t^2}{2}+\frac{t^3}{3}+o(t^3).$$

score -1 · Answer 3 · edited Sep 07 '21 at 16:18

-1

Answer seems to be $e^{1/2}. If you set the e term aside and the treat the rest as a 1^infinity form of limit you finally get e^1/2 without using any approximations

edited Sep 07 '21 at 16:18

amWhy

209,954

answered Sep 07 '21 at 14:51

Gautam Khona

1

1

Please provide additional details in your answer. As it's currently written, it's hard to understand your solution. – Community Sep 07 '21 at 15:17
1

Please add further details to expand on your answer, such as working code or documentation citations. – Community Sep 07 '21 at 15:19

$\lim_{n\rightarrow \infty}\left[\frac{\left(1+\frac{1}{n^2}\right)\cdot \cdots\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n$

3 Answers3