The question asks to find the following limit.
$$\lim_{n\to \infty}\left[\frac{\left((n+1)(n+2)...(n+n)\right)^{1/n}}{n}\right]$$
I tried taking $\log$ but I do not think that we get an L' Hopital form after that.
There is another approach that I tried. I rewrote the expression as $$\lim_{n\to \infty}\left[\left(\frac{n+1}{n}\frac{n+2}{n}...\frac{n+n}{n}\right)^{1/n}\right]$$
If I take $\log$ now, won't the limit be 0? The answer in my book is $4/e$.
Take logarithm and you get a Riemann sum for
$$\ln(S)=\int_0^1\ln(1+x)dx$$
Which on evaluation yields $$\ln(S)=2\ln(2)-1$$
Note that
$$\frac{\left((n+1)(n+2)...(n+n)\right)^{1/n}}{n}=\left[\frac{(2n)!}{ n^nn!}\right]^{\frac1n}=a_n^\frac1n$$
and by ratio-root criteria $$\frac{a_{n+1}}{a_n} \rightarrow L\implies a_n^{\frac{1}{n}} \rightarrow L$$
we have
$$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{ (n+1)^{n+1}(n+1)!}\cdot \frac{n^nn!}{(2n)!}=\frac{(2n+2)(2n+1)}{(n+1)^2}\cdot\frac{1}{\left(1+\frac1n\right)^n}\to \frac4e$$
therefore
$$\lim_{n\to \infty}\left[\frac{\left((n+1)(n+2)...(n+n)\right)^{1/n}}{n}\right]=\frac4e$$
By Riemann sum we have $$\left[\left(\frac{n+1}{n}\frac{n+2}{n}...\frac{n+n}{n}\right)^{1/n}\right] =\exp\left[\frac{1}{n}\sum_{k=1}^{n}\ln\left(1+\frac{k}{n} \right)\right] \to \exp\left[\int_0^1 \ln\left(1+x\right)dx\right] \\= \exp\left[ 2\ln2 -1\right] =\color{red}{\frac{4}{e}} $$
since $$((x+1)\ln(x+1) -(x+1))' =\ln(x+1)$$
Yet another idea: take $\log$ and apply Stolz-Cesàro: $$ \log L = \lim_{n\to\infty}\frac{\log(n+1) + \cdots + \log(n+n) - n\log n}{n} = \cdots $$
