Homework: limit of $\frac{1}{n}\sqrt[n]{\prod_{k=1}^n(n+k)}$ as $n\to\infty$

Solution 1:

You are looking for $$ \lim_{n\to +\infty}\sqrt[n]{\frac{(2n)!}{n!\cdot n^n}}. $$ By setting $a_n=\frac{(2n)!}{n!\cdot n^n}$ we have: $$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)n^n}{(n+1)(n+1)^{n+1}}=\frac{4n+2}{n+1}\cdot\frac{1}{\left(1+\frac{1}{n}\right)^n}$$ hence: $$ \lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\frac{4}{e} $$ and that implies: $$ \lim_{n\to+\infty}\sqrt[n]{\frac{(2n)!}{n!\cdot n^n}}=\color{red}{\frac{4}{e}}.$$

Solution 2:

From the idea 1, we can derive the result using the Stirling's approximation $$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} $$ hence $$\frac{1}{n}\sqrt[n]{\frac{\left(2n\right)!}{n!}}\sim\frac{1}{n}\sqrt[n]{\frac{\sqrt{2n}}{\sqrt{n}}\left(\frac{2n}{e}\right)^{2n}\left(\frac{e}{n}\right)^{n}}=\frac{2^{2+1/2n}}{e}\rightarrow\frac{4}{e}. $$

Solution 3:

Start with the expression on the right of your Idea #2. Using $\ln (n+k)=\ln n + \ln (1+k/n)$ for each $k$ shows that expression equals

$$-\ln n+\frac{1}{n}\sum_{k=1}^{n}[\ln n +\ln (1+k/n )] $$ $$= -\ln n+\frac{1}{n}[n\ln n + \sum_{k=1}^{n}\ln (1+k/n )]= \frac{1}{n}\sum_{k=1}^{n}\ln (1+k/n )$$ $$ \to \int_0^1 \ln (1+x)\, dx = 2\ln2 -1 = \ln (4/e).$$

Exponentiating back gives $4/e$ for the limit.