A limit of a product via Riemann sums

Step 1: take logarithms. Let
$$u_n=\frac{1}{n}\sqrt[n]{(n+1)(n+2)\cdots(2n)}>0.$$
Then
$$\ln(u_n)=-\ln n+\frac{1}{n}\sum_{k=1}^{n}\ln(n+k).$$
Factor $n$ in each factor inside the sum:
$$\ln(n+k)=\ln n+\ln\!\left(1+\frac{k}{n}\right).$$
Therefore
$$\frac{1}{n}\sum_{k=1}^{n}\ln(n+k)=\frac{1}{n}\sum_{k=1}^{n}\left(\ln n+\ln\!\left(1+\frac{k}{n}\right)\right)=\ln n+\frac{1}{n}\sum_{k=1}^{n}\ln\!\left(1+\frac{k}{n}\right),$$
since the constant $\ln n$ is summed $n$ times and then divided by $n$. Hence
$$\ln(u_n)=\frac{1}{n}\sum_{k=1}^{n}\ln\!\left(1+\frac{k}{n}\right).$$

Step 2: Riemann sum. The expression
$$\frac{1}{n}\sum_{k=1}^{n}\ln\!\left(1+\frac{k}{n}\right)$$
is a left Riemann sum for the continuous function $t\mapsto\ln(1+t)$ on $[0,1]$: the sample points are $t_k=k/n$ with step $1/n$, for $k=1,\ldots,n$, i.e. on the partition $[0,1/n],[1/n,2/n],\ldots,[(n-1)/n,1]$. As $n\to+\infty$,
$$\frac{1}{n}\sum_{k=1}^{n}\ln\!\left(1+\frac{k}{n}\right)\longrightarrow\int_0^1\ln(1+t)\,dt.$$

Step 3: compute the integral. Substituting $s=1+t$, $ds=dt$, with $s$ from $1$ to $2$:
$$\int_0^1\ln(1+t)\,dt=\int_1^2\ln s\,ds.$$
An antiderivative of $\ln s$ is $s\ln s-s$; hence
$$\int_1^2\ln s\,ds=\bigl[s\ln s-s\bigr]_1^2=(2\ln2-2)-(0-1)=2\ln2-1.$$
So
$$\ln(u_n)\longrightarrow 2\ln 2-1=\ln 4-\ln e=\ln\!\left(\frac{4}{e}\right).$$

Step 4: exponentiate. By continuity of $\exp$,
$$u_n=\exp(\ln u_n)\longrightarrow\exp\!\left(\ln\frac{4}{e}\right)=\frac{4}{e}.$$