Stirling's Approximation and Ramanujan's Proof of Bertrand's Postulate Perhaps, the most astounding step in Ramanujan's proof of Betrand's postulate is his application of Stirling's approximation.
He starts with the following inequality:
$\log\Gamma(x) - 2\log\Gamma(\frac{1}{2}x + \frac{1}{2}) \le \log[x]! - 2\log[\frac{1}{2}x]! \le \log\Gamma(x+1) - 2\log\Gamma(\frac{1}{2}x + \frac{1}{2})$
Then, applying Stirling's approximation, Ramanujan gets to:
$\log[x]! - 2\log[\frac{1}{2}x]! < \frac{3}{4}x$ if $x > 0$
and
$\log[x]! - 2\log[\frac{1}{2}x]! > \frac{2}{3}x$ if $x > 300$
I would be very interested in understanding how Stirling's approximation gets us to these two conclusions.
As I understand it, Ramanujan is refering to Stirling's Approximation for the Gamma function which as I understand to be this (from Wikipedia):
$\Gamma(z) = \sqrt{\frac{2\pi}{z}}(\frac{z}{e})^{z}(1 + O(\frac{1}{z}))$
If someone could provide the details, I would greatly appreciate it!  :-)
 A: To make the bounds explicit we can start with the earlier form of Stirling's Approximation
$$
\log \Gamma(z) = \left(z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log (2\pi)
+ \sum_{n=1}^{k-1} \frac{B_{2n}}{2n(2n-1)z^{2n-1}}+R_k(z)
$$
 If we take $k-1$ terms in the sum then $|R_k|$ is bounded by the $k$th term (see the answers to this question and the citations). In particular take $k=1$ then $|R_1|\le\frac{1}{12z}$.
Then
$$
\begin{align}
\log \Gamma(z)-2\log\Gamma\left(\frac{z+1}{2}\right) 
&= z\log2-z\log\frac{z+1}{z}-\frac{1}{2}\log\frac{2\pi z}{e^2}+R_1(z)-2R_1\left(\frac{z+1}{2}\right)\\
&=z\log2+\Delta_1(z) \\
&=\frac{2}{3}z + (Az + \Delta_1(z))
\end{align}
$$
where $A=\log 2-2/3=0.02648\cdots$. Now we need to find the minimum $z$ that guarantees $Az+\Delta_1(z)>0$. Using $|R_1(z)-2R_1((z+1)/2)|< \frac{5}{12z}$ numerically we find $z>126$ is sufficient.
For the other side
$$
\begin{align}
\log \Gamma(z+1)-2\log\Gamma\left(\frac{z+1}{2}\right) 
&= z\log2+\frac{1}{2}\log\frac{z+1}{2\pi}+R_1(z+1)-2R_1\left(\frac{z+1}{2}\right)\\
&=z\log2+\Delta_2(z) \\
&=\frac{3}{4}z - (Bz - \Delta_2(z))
\end{align}
$$
where $B=3/4-\log 2=0.05685\cdots$. Using $|R_1(z+1)-2R_1((z+1)/2)|\le \frac{5}{12(z+1)}$ we find that $Bz-\Delta_2(z)>0$ for all $z>0$.
