Prove that $1+\frac{1}{8n}\sqrt{\pi_1n}<\frac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<(1+\frac{1}{8n}+\frac{1}{128n^{2}})\sqrt{\pi_2n}$ Within the confines of the O-level syllabus, prove that:
$$1+\dfrac{1}{8n}\sqrt{\pi_1n}<\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}$$
for all positive integer $n$, where $\pi_1=3.141$ and $\pi_2=3.142$
Using A-level mathematics, show that the result remains true with $\pi_1=\pi_2=\pi$ where $\pi$ is the familiar constant associated with a circle.
This is a question found in Hammersley's "On the enfeeblement of mathematical skills by "Modern Mathematics" and by similar soft intellectual trash in schools and universities"
When I first look at this problem, I wonder if I can prove it via Wallis'product?
$$\prod_{i=1}^{\infty}\dfrac{2n}{2n-1}.\dfrac{2n}{2n+1}=\dfrac{\pi}{2}$$
 A: $$
\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)} = \dfrac{2n!}{(2n-1)!} = 2n \\ \therefore \;\; 1+\dfrac{1}{8n}\sqrt{\pi_1n}<\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
\\ = 1+\dfrac{1}{8n}\sqrt{\pi_1n}<2n<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
$$
Considering the case of $$\pi_1 = \pi_2 = \pi$$ since this generality can be used for the other case too. 
Splitting the inequalities into two, 
\begin{align}
I_1: 
1+\dfrac{1}{8n}\sqrt{\pi n}<2n \\
1+\dfrac{1}{8\sqrt n}\sqrt{\pi}<2n \\
\dfrac{1}{8\sqrt n}\sqrt{\pi}<2n - 1 \\
\dfrac{1}{8}\sqrt{\pi}<(2n - 1)\cdot\sqrt n \tag{1}\label{1}
\end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in \eqref{1} gives $$ 0.2215 < 1 $$ which is true. 
\begin{align}
I_2:
2n<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi n}\\
\dfrac{2n}{\sqrt{\pi n}}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1<\left(\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1<\dfrac{1}{8n}\left(1+\dfrac{1}{16n}\right)\\
\dfrac{2n}{\sqrt{\pi n}} - 1<\dfrac{1}{8n}\left(\dfrac{16n+1}{16n}\right)\\
(128n^{2})\cdot\dfrac{2\sqrt n}{\sqrt{\pi}} - 1< 16n+1\\
\dfrac{256n^{\frac{5}{2}}}{\sqrt \pi} - 128n^{2}<16n+1\\
\dfrac{256}{\sqrt \pi} < ( 128n^2 + 16n + 1 )\cdot n^{\frac{2}{5}} \tag{2}\label{2}
\end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in \eqref{2} gives $$ 144.43 < 145 $$ which is true. 
