Prove limit $\lim\limits_{n\to\infty} \frac{2^{2n+1} \cdot r^{2n} \pi^n \cdot n!}{(2n+1)!} = 0$ I would like to prove that
$\lim\limits_{n\to\infty}\dfrac{2^{2n+1} \cdot r^{2n} \pi^n \cdot n!}{(2n+1)!}=0$
I have thought of De L'Hospital but that would require me to differentiate the gamma function which doesn't seem very helpful.
 A: Hint
Use the following inequality,
$$
{
(2n+1)!\ge n!\cdot n!,
}
$$
and use $\lim_{n\to\infty} \frac{a^n}{n!}=0$ for any $a\in \Bbb R$.
A: By ratio test
$$\frac{2^{2n+3} \cdot r^{2n+2} \pi^{n+1} \cdot (n+1)!}{(2n+3)!} \frac{(2n+1)!}{2^{2n+1} \cdot r^{2n} \pi^n \cdot n!}=\frac{2^2\cdot r^2\pi\cdot (n+1)}{(2n+3)(2n+2)}$$
A: We want to get the limit :
$$ L=\lim\limits_{n\to\infty} \frac{2^{2n+1} \cdot r^{2n} \pi^n \cdot n!}{(2n+1)!} = 0 $$
Intuitively , we can check that :
(A) $ 2^{2n+1}=2 \times {(2^2)}^{n}=2 \times {4}^{n} $
(B) $ r^{2n}={(r^2)}^{n} $
(C) $ (2n+1)!=(n!)(n+1 \times \cdots \times 2n)(2n+1) $
Thus we get :
$$ L=2\lim\limits_{n\to\infty} \frac{ ( 4 \cdot (r^2) \cdot \pi ) ^{n} \cdot n!}{(n!)(n+1 \times \cdots \times 2n)(2n+1)} $$
$$ L=2\lim\limits_{n\to\infty} \frac{ ( K ) ^{n}}{(n+1 \times \cdots \times 2n)} \times \frac{n!}{n!} \times \frac{1}{(2n+1)} $$
$$ L=2\lim\limits_{n\to\infty} X \times \frac{1}{(2n+1)} $$
Here , $X$ has $n$ terms in Numerator with $n$ terms in Denominator.
More-over , Numerator has constant terms while Denominator has infinitely larger terms.
Hence , limit $X$ can not exceed $1$ [[ $X$ has limit Exactly $0$ , but we do not require that limit ]] , while the last term goes to $0$.
Hence , limit $ L=2 \times X \times 0 = 0 $
A: Let $\;x_n=\dfrac{2^{2n+1}\cdot r^{2n}\pi^n\cdot n!}{(2n+1)!}\;$ for any $\;n\in\Bbb N\;$ where $\;r>0\,.$
It results that
$\begin{align}0<x_{n+1}&=\dfrac{2^{2n+3}\cdot r^{2n+2}\pi^{n+1}\cdot(n+1)!}{(2n+3)!}=\dfrac{4\pi r^2(n+1)}{(2n+3)(2n+2)}\!\cdot\!x_n=\\&=\dfrac{2\pi r^2}{2n+3}\!\cdot\!x_n<\dfrac{\pi r^2}n\!\cdot\!x_n<x_n\quad\text{ for any }n>\lfloor\pi r^2\rfloor\,.\end{align}$
Hence, the sequence $\,\{x_n\}\,$ is positive and eventually decreasing, consequently, there exists $\;\lim\limits_{n\to\infty}x_n=l\in\Bbb R^+_0\,.$
Since $\;x_{n+1}=\dfrac{2\pi r^2}{2n+3}\!\cdot\!x_n\;$ for any $\;n\in\Bbb N\;,\;$ we get that
$\lim\limits_{n\to\infty} x_{n+1}=\lim\limits_{n\to\infty}\dfrac{2\pi r^2}{2n+3}\cdot\lim\limits_{n\to\infty}x_n\;\;,\;\;$ that is ,
$l=0\!\cdot\!l=0\;.$
So we have proved that
$\lim\limits_{n\to\infty}x_n=0\;.$
A: Consider the function$$f(n)= \frac{2^{2n+1} \, r^{2n}\, \pi^n \, n!}{(2n+1)!} $$
Taking logarithms, using Stirling approximation and finishing with Taylor series for large values of $n$, we have
$$\log[f(n)]=n (-\log (n)+2 \log (r)+1+\log (\pi ))-\frac{1}{2} (2 \log
   (n)+\log (2))+O\left(\frac{1}{n}\right)$$ Its derivative
$$\left(\log[f(n)] \right)'=(-\log (n)+2 \log (r)+\log (\pi
   ))-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ cancels at
$$n_*=-\frac{1}{W\left(-\frac{1}{\pi  r^2}\right)}$$ and the second derivative test shows that this is a maximum since
$$\left(\log[f(n)] \right)''=-\frac 1n+\frac 1 {n^2}+O\left(\frac{1}{n^3}\right)$$
So $f(n)$ starts at $\frac{4 \pi  r^2}{3}$, goes through a maximum value  $f(n_*)$ and decreases to $0$.
Its asymptotics is
$$f(n) \sim \frac 1{\sqrt 2}\left(\frac{e \pi r^2}{n}\right)^n\,\exp\left(-\frac{11}{24 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right) \right)$$
Trying for $r=1$, the approximation gives $n_*=1.80562$ and $f(n_*)=5.22077$ while the exact values are $n_*=2.12847$ and $f(n_*)=5.27777$.
The ratio of $f(n)$ to its asymptotics is larger than $0.999$ as soon as $n \geq 4$ and larger than $0.9999$ as soon as $n \geq 8$.
