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

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 $n \to \infty$.