Evaluate $\lim_{n \rightarrow \infty} \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$

By a Riemann sum argument, $$ \log\frac{\left[(n+1)(n+2)\cdot\ldots\cdot(n+n)\right]^{\frac{1}{n}}}{n}=\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)$$ converges towards: $$ \int_{0}^{1}\log(1+x)\,dx = -1+\log 4,$$ hence the value of the limit is $\large\color{red}{\frac{4}{e}}$.

You can't apply the theorem that $a_n \to l$ implies $b_n = (a_1a_2\cdots a_n)^{1/n}$ to this limit since the $k$-th term, i.e. $1+\tfrac{k}{n}$, depends on $n$.

If you take the natural log of the limitand(?) you get:

$\ln\left[\left(\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)\cdots\left(1+\dfrac{n}{n}\right)\right)^{1/n}\right] = \dfrac{1}{n}\displaystyle\sum_{k = 1}^{n}\ln\left(1+\dfrac{k}{n}\right)$,

which is a Riemann sum for $\displaystyle\int_{0}^{1}\ln(1+x)\,dx$.

So, as $n \to \infty$, we have $\dfrac{1}{n}\displaystyle\sum_{k = 1}^{n}\ln\left(1+\dfrac{k}{n}\right) \to \displaystyle\int_{0}^{1}\ln(1+x)\,dx = 2\ln 2 - 1$.

Exponentiating both sides gives us $(1+\tfrac{1}{n})(1+\tfrac{2}{n})\cdots(1+\tfrac{n}{n}))^{1/n} \to e^{2\ln 2 - 1} = \dfrac{4}{e}$ as $n \to \infty$.

Hint: You may use $$\lim\limits_{n \to \infty} \sqrt[n]{a_n} = \lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}$$

with $a_n = \dfrac {(n+1)(n+2)\cdots(n+n)}{n^n}$

The Riemann integral way is nice too, but if you insist on taking $\log$ you could apply Stolz-Cesaro Theorem too:

$$\lim\limits_{n \to \infty} \frac{-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)}{n} \\= \lim\limits_{n \to \infty} \left(-(n+1)\log (n+1) + \sum\limits_{k=1}^{n+1} \log \left(k+n\right)\right) - \left(-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)\right) \\ = \lim\limits_{n \to \infty} \log 2 + \log \frac{2n+1}{n+1} - n\log \left(1+\frac{1}{n}\right) = \log 4 - 1$$

Giving you the desired limit $\dfrac{4}{e}$.