Evaluate $\lim_{n \rightarrow \infty} \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$ Evaluate $$\lim_{n \rightarrow \infty~} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$
Attempt: Let $$y=\lim_{n \rightarrow \infty} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$
$$\implies \log y = \lim_{n \rightarrow \infty} \dfrac {1} {n} [\log (n+1) +\cdots+log(n+n)-log(n)] $$
How do I move forward?
Thank you very much for your help.
 A: 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 $\dfrac{4}{e}$.
A: Your mistake is
$$ \dfrac {1} {n} [\log (n+1) +\cdots+\log(n+n)-\color{red}{n}\log(n)] =\frac1n\sum_{k=1}^n\log\left(1+\frac kn\right)$$
and use the Riemann sum.
A: Let $$A_n=\frac 1n \left(\prod_{i=1}^n (n+i)\right)^{\frac 1n}$$ Using Pochhammer notation and using the corresponding expression using the gamma function, we have $$\prod_{i=1}^n (n+i)=(n+1)_n=\frac{2^{2 n} }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)$$ So, taking logarithms $$\log(A_n)=-\log(n)+\frac 1n\log \left(\frac{2^{2 n} \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}\right)$$ Now, using Stirling approximation for the gamma function, we have $$\log \left(\frac{2^{2 n} \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}\right)=n (\log (n)+2 \log (2)-1)+\frac{\log (2)}{2}-\frac{1}{24
   n}+O\left(\frac{1}{n^{5/2}}\right)$$ which makes $$\log(A_n)=\left(2 \log (2)-1\right)+\frac{\log (2)}{2
   n}-\frac{1}{24 n^2}+O\left(\frac{1}{n^{7/2}}\right)$$ and then the limit $\frac 4e$.
Edit
For sure, using Riemann sum as JimmyK4542 just answered makes the problem simpler. The approach I proposed shows, beside the limit itself, how it is approached.
A: For some reason nobody mentioned Stirling's approximation:
$$
\frac{(2n)!^{1/n}}{n!^{1/n} n} \sim \frac{(\sqrt{4\pi n}(2n/e)^{2n})^{1/n}}{(\sqrt{2\pi n}(n/e)^n)^{1/n} n} \sim \frac{(2n/e)^2}{(n/e)n} = \frac{4}{e}.
$$
A: You can try this easy one:
$\lim_{n\to \infty}\dfrac{1}{n}{[(n+1)(n+2)...(n+n)]}^{\frac{1}{n}}=\lim _{n\to \infty}\dfrac{1}{n}[n^n(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n})]^{\frac{1}{n}}$
$=\lim_{n\to \infty}{\prod_{k=1}^n}(1+\frac{k}{n})^{\frac{1}{n}}$
Let $y={\prod_{k=1}^n}(1+\frac{k}{n})^{\frac{1}{n}}\implies \ln y=\frac{1}{n}\sum _{k=1}^n\ln (1+\frac{k}{n})=\int _0^ 1\ln(1+x)\operatorname{dx}$
$\implies y=\frac{4}{e}$
A: Let $A_n$ be the given expression, then: $A_n = \displaystyle \prod_{k=1}^n\left(1+\dfrac{k}{n}\right)^{\frac{1}{n}}\Rightarrow \ln A_n =\dfrac{1}{n}\cdot \displaystyle \sum_{k=1}^n \ln\left(1+\dfrac{k}{n}\right)\Rightarrow \displaystyle \lim_{n\to \infty}A_n = \displaystyle \int_{0}^1 \ln(1+x)dx= ...$
A: 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 $\large\color{red}{\frac{4}{e}}$.
A: 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 $n \to \infty$.
