<s>[INST] Find the integral part of the product $\frac{2}{1} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7} \cdots \frac{2016}{2015}.$ 
Find the integral part of the following number $$T = \dfrac{2}{1} \cdot \dfrac{4}{3} \cdot \dfrac{6}{5} \cdot \dfrac{8}{7} \cdots \dfrac{2016}{2015}.$$

We can show that $T = 2017\int_{0}^{\frac{\pi}{2}} \sin^{2017}(x)dx$, since $$\int_0^{\frac{\pi}{2}} \sin^{2n+1}(x) dx = \dfrac{2}{3} \cdot \dfrac{4}{5} \cdot \dfrac{6}{7} \cdots \dfrac{2n}{2n+1},$$ but how do we calculate the integral part of $2017\int_{0}^{\frac{\pi}{2}} \sin^{2017}(x)dx$?
 [/INST] We may prove the inequality mentioned by achille hui in the comments without resorting to Stirling's approximation. For large values of $n$, we have:
$$ \frac{(2n)!!}{(2n-1)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^{-1} \tag{1} =2\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^{-1}$$
and since $\left(1-\frac{1}{2k}\right)^{2}$ is close to $\left(1-\frac{1}{k}\right)$, that is the term of a telescopic product,
$$ \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 = 4\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)^{-1} \prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)\tag{2} $$
and since $\prod_{k\geq 2}\left(1-\frac{1}{(2k-1)^2}\right)=\frac{\pi}{4}$ by Wallis product,
$$ \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 = \pi n \prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)\tag{3} $$
where:
$$\prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)\approx \exp\sum_{k>n}\frac{1}{4k(k-1)} = \exp\left(\frac{1}{4n}\right) \tag{4}$$
by "telescopic luck" again. Here $\approx$ is actually a $\leq$. With similar arguments one may prove
$$ \exp\left(\frac{1}{8n}-\frac{1}{96n^3}\right)\leq\frac{1}{\sqrt{\pi n}}\cdot\frac{(2n)!!}{(2n-1)!!}\leq \exp\left(\frac{1}{8n}\right) \tag{5}$$
and since $\sqrt{1008\cdot \pi}=56.273\ldots$, $\color{blue}{\large 56}$ is the correct answer.
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