evaluate: $ \lim_{n→∞} \frac1n ((n+1)(n+2)(n+3)⋯(2n))^{\frac1n}$. we have
$$
((n+1)(n+2)(n+3)⋯(2n))=\frac{(2n)!}{n!}
$$
Using Stirling formula
https://en.wikipedia.org/wiki/Stirling%27s_approximation
$$
\log_e(n!)≈n\log_en−n
$$
 in
$$
y=\frac1n((2n)!/n!)^{1/n}
$$
we have
$$
y=\frac1n\left(\frac{(2n)^{2n} e ^{−2n}}{n^n e^{−n}}\right)^{1/n}=\frac4e
$$
So the answer is
$$
\lim_{n→∞} \frac1n ((n+1)(n+2)(n+3)⋯(2n))^{\frac1n}=\frac4e
$$
Here the question is: $$\lim_{n\to\infty}\frac{1}{n}\Big((n+1)(n+2)\cdots(2n)\Big)^{\frac 1n}~?$$
 A: $$L=\lim_{n\rightarrow \infty}\frac{1}{n} \left ((n+1) (n+2)(n+3)....(2n) \right)^{1/n}$$ $$L=\lim_{n \rightarrow \infty} \left( (1+1/n) (1+2/n) (1+3/n) (1+4/n).....\right)^{1/n}$$
$$\ln L= \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\,\ln\left(1+\frac{k}{n} \right)= \int_{0}^{1} \ln (1+x) dx=(1+x)\ln (1+x)-(1+x)|_{0}^{1}=\ln (4/e)$$ $$\Rightarrow L= 4/e. $$
A: To get even a bit more than the limit.
$$y=\frac 1 n \frac{(2 n)!}{n!}\implies \log(y)=\log((2n)!)-\log(n!)-\log(n)$$ Using one more term in Stirling approximation and simplifying
$$\log(y)=(2 \log (2)-1)+\frac{\log (2)}{2 n}+O\left(\frac{1}{n^2}\right)$$ Continuing with Taylor series
$$y=e^{\log(y)}=\frac{4}{e}+\frac{2 \log (2)}{e n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
Using your pocket calculator, try for $n=10$ (far away from $\infty$ !). The exact result would be $1.52278$ while the above truncated series would give $1.52252$.
A: If
$\begin{array}\\
f(n)
&=((n+1)(n+2)(n+3)⋯(2n))\\
&=\prod_{k=1}^n(n+k)\\
g(n)
&=\ln(f(n))\\
&=\sum_{k=1}^n\ln(n+k)\\
&=\sum_{k=1}^n(\ln(n)+\ln(1+k/n)\\
&=n\ln(n)+\sum_{k=1}^n\ln(1+k/n)\\
\text{so}\\
\dfrac{g(n)}{n}
&=\ln(n)+\frac1{n}\sum_{k=1}^n\ln(1+k/n)\\
\text{or}\\
\dfrac{g(n)}{n}-\ln(n)
&=\frac1{n}\sum_{k=1}^n\ln(1+k/n)\\
&\to\int_0^1 \ln(1+x)dx
\qquad\text{Riemann sum approximation}\\
&\int_1^2 \ln(x)dx\\
&=(x\ln(x)-x)|_1^2\\
&=(2\ln(2)-2)-(-1)\\
&=\ln(4)-1\\
&=\ln(4/e)\\
\end{array}
$
