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:

$$1+\dfrac{1}{8n}\sqrt{\pi_1n}&lt;\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}$$

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?
$$\prod_{i=1}^{\infty}\dfrac{2n}{2n-1}.\dfrac{2n}{2n+1}=\dfrac{\pi}{2}$$
$$
\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)} = \dfrac{2n!}{(2n-1)!} = 2n \\ \therefore \;\; 1+\dfrac{1}{8n}\sqrt{\pi_1n}&lt;\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
\\ = 1+\dfrac{1}{8n}\sqrt{\pi_1n}&lt;2n&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
$$

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}&lt;2n \\
1+\dfrac{1}{8\sqrt n}\sqrt{\pi}&lt;2n \\
\dfrac{1}{8\sqrt n}\sqrt{\pi}&lt;2n - 1 \\
\dfrac{1}{8}\sqrt{\pi}&lt;(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 $$ 0.2215 &lt; 1 $$ which is true. 

\begin{align}
I_2:
2n&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi n}\\
\dfrac{2n}{\sqrt{\pi n}}&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1&lt;\left(\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1&lt;\dfrac{1}{8n}\left(1+\dfrac{1}{16n}\right)\\
\dfrac{2n}{\sqrt{\pi n}} - 1&lt;\dfrac{1}{8n}\left(\dfrac{16n+1}{16n}\right)\\
(128n^{2})\cdot\dfrac{2\sqrt n}{\sqrt{\pi}} - 1&lt; 16n+1\\
\dfrac{256n^{\frac{5}{2}}}{\sqrt \pi} - 128n^{2}&lt;16n+1\\
\dfrac{256}{\sqrt \pi} &lt; ( 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 $$ 144.43 &lt; 145 $$ which is true.
$$
\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)} = \dfrac{2n!}{(2n-1)!} = 2n \\ \therefore \;\; 1+\dfrac{1}{8n}\sqrt{\pi_1n}&lt;\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
\\ = 1+\dfrac{1}{8n}\sqrt{\pi_1n}&lt;2n&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
$$

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}&lt;2n \\
1+\dfrac{1}{8\sqrt n}\sqrt{\pi}&lt;2n \\
\dfrac{1}{8\sqrt n}\sqrt{\pi}&lt;2n - 1 \\
\dfrac{1}{8}\sqrt{\pi}&lt;(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 $$ 0.2215 &lt; 1 $$ which is true. 

\begin{align}
I_2:
2n&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi n}\\
\dfrac{2n}{\sqrt{\pi n}}&lt;\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1&lt;\left(\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1&lt;\dfrac{1}{8n}\left(1+\dfrac{1}{16n}\right)\\
\dfrac{2n}{\sqrt{\pi n}} - 1&lt;\dfrac{1}{8n}\left(\dfrac{16n+1}{16n}\right)\\
(128n^{2})\cdot\dfrac{2\sqrt n}{\sqrt{\pi}} - 1&lt; 16n+1\\
\dfrac{256n^{\frac{5}{2}}}{\sqrt \pi} - 128n^{2}&lt;16n+1\\
\dfrac{256}{\sqrt \pi} &lt; ( 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 $$ 144.43 &lt; 145 $$ which is true.