$ \lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\frac{1}{\sqrt{n}}$ 
Knowing that : 
  $$I_n=\int_0^{\frac{\pi}{2}}\cos^n(t) \, dt$$
$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1$$
  $$I_{n}\sim \sqrt{\dfrac{\pi}{2n}}$$
Calculate: 
  $$
\lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\dfrac{1}{\sqrt{n}}$$

Indeed,
$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1 \\
\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}=\dfrac{2}{\pi}\times I_{2n}$$
then
$$
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&=\dfrac{2}{\pi}\times I_{2n}\times (2n+1)\times\dfrac{1}{\sqrt{n}}\\
&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{\sqrt{n}} \times \sqrt{ \dfrac{2n\times I^{2}_{2n}}{2n}  } \\
&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ 2n\times I^{2}_{2n}   } \\
\end{align*}
$$
or $$(2n)I^{2}_{2n}\sim \dfrac{\pi}{2}$$
then
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ 2n\times I^{2}_{2n}   } \\
&\sim\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times  \sqrt{ \dfrac{\pi}{2}   } \\
\end{align*}
$$
I'm stuck here 
i think that i can go ahead 
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&\sim \dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times  \sqrt{ \dfrac{\pi}{2}   } \\
&\sim  \dfrac{2}{\pi}\times\dfrac{2n}{ \sqrt{2}\times n} \times  \sqrt{ \dfrac{\pi}{2}   } \\
&\sim \frac{ 2\sqrt{\pi} }{\pi}=\dfrac{2}{\sqrt{\pi}}
\end{align*}
am i right ? if that so is there any other way
 A: So we know that:
$$ I_{2n}=\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{\pi}{2}\cdot\frac{(2n-1)!!}{(2n)!!}=\frac{\pi}{2}\cdot\frac{1}{4^n}\binom{2n}{n}.\tag{1}$$
If, in the same way, we prove:
$$ I_{2n+1}=\int_{0}^{\pi/2}\cos^{2n+1}(\theta)\,d\theta = \frac{(2n)!!}{(2n+1)!!}\tag{2} $$
then we have:
$$\lim_{n\to +\infty} n\,I_{2n}\, I_{2n+1} = \lim_{n\to +\infty}\frac{\pi n}{4n+2}=\frac{\pi}{4}.\tag{3}$$
Since $\{I_n\}_{n\in\mathbb{N}}$ is a decreasing sequence, we have $\frac{I_{2n+1}}{I_{2n}}\leq 1$ as well as:
$$ \frac{I_{2n+1}}{I_{2n}}\geq \frac{I_{2n+2}}{I_{2n}} = \frac{2n+1}{2n+2}\geq 1-\frac{1}{2n}\tag{4}$$
hence, by squeezing:
$$ \lim_{n\to +\infty} n I_{2n}^2 = \frac{\pi}{4}, \tag{5}$$
then:

$$ \lim_{n\to +\infty}\frac{(2n+1)!!}{(2n)!!}\cdot\frac{1}{\sqrt{n}}=\color{red}{\frac{2}{\sqrt{\pi}}}.\tag{6}$$

A: We are given 
$$I_{2n}=\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!} \tag 1$$
and
$$I_n\sim \sqrt{\frac{\pi}{2n}} \tag 2$$
From $(2)$ is trivial to see that
$$I_{2n}\sim \frac12 \sqrt{\frac{\pi}{n}} \tag 3$$
Then, using $(1)$ and $(3)$, we find that 
$$\begin{align}
\frac{(2n+1)!!}{\sqrt{n}(2n)!!}&=\frac{(2n+1)(2n-1)!!}{\sqrt{n}(2n)!!}\\\\
&=\frac{2n+1}{\sqrt{n}}\left(\frac{(2n-1)!!}{(2n)!!}\right)\\\\
&=\frac{2n+1}{\sqrt{n}}\left(\frac{2}{\pi}I_{2n}\right)\\\\
&\sim \frac{2n+1}{\sqrt{n}}\left(\frac{2}{\pi}\frac12\sqrt{\frac{\pi}{n}}\right)\\\\
&=\frac{2n+1}{n\sqrt{\pi}}\\\\
&\to \frac{2}{\sqrt{\pi}}\,\,\text{as}\,\,n\to \infty
\end{align}$$
Thus, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{(2n+1)!!}{\sqrt{n}(2n)!!}=\frac{2}{\sqrt{\pi}}}$$
