If

$f\left( n \right) =\frac { 1 }{ n } { \big\{ (2n+1)(2n+2)\ldots(2n+n) \big\} }^{ 1/n },$

then what is $\displaystyle \lim _{ n\rightarrow \infty }{ f\left( n \right) }?$

$\frac 4e$
$\frac{27}{4e}$
$4e$
$\frac{27e}4$

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We have $f(n)=\prod_{k=1}^{n}\left(2+\frac{k}{n}\right)^{1/n}$ , with $\ln(f(n))$ $=\sum_{k=1}^{n}\ln{\left(2+\frac{k}{n}\right)}\frac{1}{n}$ , a (right) Riemann sum of $\ln(x)$ between 2 and 3. Thus $\lim_{n\to\infty}\ln(f(n))$ $=\int_{2}^{3}\ln(x)dx=[x\ln(x)-x]_{2}^{3}$ $=3\ln(3)-2\ln(2)-1$ . Now $\lim_{n\to\infty}f(n)=e^{\lim_{n\to\infty}\ln(f(n))}=\frac{3^3}{2^2e}=\boxed{\frac{27}{4e}},$ by the continuity of $e^x$ .