Find the value of $\,\, \lim_{n \to \infty}\Big(\!\big(1+\frac{1}{n}\big)\big(1+\frac{2}{n}\big) \cdots\big(1+\frac{n}{n}\big)\!\Big)^{\!1/n} $

Question

What is the limit of:

$$ \lim_{n \to \infty}\bigg(\Big(1+\frac{1}{n}\Big)\Big(1+\frac{2}{n}\Big) \cdots\Big(1+\frac{n}{n}\Big)\bigg)^{1/n}? $$

By computer, I guess the limit is equal to $\dfrac{4}{e}$, but I have no idea about proving that.

Thank you for your any help.

Answer 1

Let $f(n)=[\prod_{i=0}^n (1+r/n)]^{1/n}$ Then, $\ln f(n)=1/n\sum_{r=0}^n\ln (1+r/n)\Rightarrow \lim_{n\to \infty}\ln f(n)=\int_{0}^1 \ln (x+1) dx=\ln 2-\int_{0}^1 \dfrac{x}{x+1}dx=2\ln 2-1\Rightarrow \lim_{n\to \infty} f(n) =4/e$

Answer 2

We shall use that:

Fact. $\displaystyle\lim_{n\to\infty}\left(\frac{n^n}{n!}\right)^{\!1/n} \!\!\!\!\!\longrightarrow\mathrm{e}.$

Then $$\bigg(\Big(1+\frac{1}{n}\Big)\Big(1+\frac{2}{n}\Big) \cdots\Big(1+\frac{n}{n}\Big)\bigg)^{1/n}= \left(\frac{(2n)!}{n^n n!}\right)^{1/n}=\left(\frac{2^{2n}\frac{n^n}{n!}}{\frac{(2n)^{2n}}{(2n)!}}\right)^{\!1/n}=4\frac{\left(\frac{n^n}{n!}\right)^{1/n}}{\left(\frac{(2n)^{2n}}{(2n)!}\right)^{1/n}} \longrightarrow\frac{4\mathrm{e}}{\mathrm{e}^2}=\frac{4}{\mathrm{e}}. $$

Note. The Fact above can be shown as follows: $$ \log \left(\frac{n!}{n^n}\right)^{1/n}=\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right)\to\int_0^1\log x\,dx=-1. $$