Evaluate $\lim_{n\rightarrow\infty}$ $\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$

Question

Question: Evaluate lim$_{n\rightarrow\infty}$$\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$

My Approach Let $a_{n}=\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}$

$$a_{n+1}=\frac{\left(n+2\right)\left(n+3\right).....\left(2n+2\right)}{\left(n+1\right)^{\left(n+1\right)}}$$

$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}=\frac{4}{\exp (1)}$$

Using Cauchy's second theorem On Limits

Lim$_{n\rightarrow\infty}\left[\frac{a_{2}}{a_{1}}\frac{a_{3}}{a_{2}}..........\frac{a_{n+1}}{a_{n}}\right]^{\frac{1}{n}}$=Lim$_{n\rightarrow\infty}$$\left[\frac{a_{n+1}}{a_{1}}\right]^{\frac{1}{n}}=\frac{4}{\exp(1)}$

Now I don't know how to move forward

Answer 1

$$\begin{eqnarray*}\lim_{n\to +\infty}\left[\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)\right]^{1/n}&=&\exp\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)\\&=&\exp\int_{0}^{1}\log(1+x)\,dx=\color{red}{\frac{4}{e}}.\end{eqnarray*} $$

Answer 2

Let the limit be $L$ then:

$$\begin{align} \ln(L) &= \lim_{n\to \infty}\frac{1}{n} \sum_{i=1}^{n} \ln\left( 1 + \frac{i}{n} \right) \\ &= \int_{0}^{1}\ln( 1+x ) dx \\ &= (x+1)\ln(1+x) -x |_{0}^{1} \\ &= 2\ln(2) - 1 \end{align}$$

So the required limit is $\exp({2\ln (2) -1}) = \color{blue}{\dfrac{4}{e}}$