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 $x > 0$


$\log[x]! - 2\log[\frac{1}{2}x]! > \frac{2}{3}x$ 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! :-)

  • 2
    $\begingroup$ Taking logarithms of the approximation you wrote down yields $\log\Gamma(z) = \frac12\log\frac{2\pi}z + z\log\frac ze + O(\frac1z)$. If you plug in $z$ and $\frac z2$, you'll get an asymptotic formula for $\log\Gamma(z)-2\log\Gamma(\frac z2)$ that should be of help to you. $\endgroup$ – Greg Martin Mar 27 '13 at 8:03
  • $\begingroup$ @Greg, Thanks for the tip. So, this will help: $\log\Gamma(z) - 2\log\Gamma(\frac{z}{2}) \sim \frac{1}{2}\log\frac{2\pi}{z} + z\log\frac{z}{e} - \log\frac{4\pi}{z} - z\log\frac{z}{2e}$ $\endgroup$ – Larry Freeman Mar 28 '13 at 4:55
  • 1
    $\begingroup$ Yes, which simplifies to $z\log2 + a\log z + b + O(1/z)$ for some constants $a$ and $b$. $\endgroup$ – Greg Martin Mar 28 '13 at 5:45
  • $\begingroup$ ok. So, proving $\log[x]! - 2\log[\frac{1}{2}x]! > \frac{2}{3}x$ if $x > 300$ consists of showing that $\log\Gamma(301) - 2\log\Gamma(\frac{301}{2}) > (\frac{2}{3})*{301}$ and $\frac{d}{dx}(\log\Gamma(x) - 2\log\Gamma(\frac{1}{2}x)) > \frac{2}{3}$. Is that right? $\endgroup$ – Larry Freeman Mar 28 '13 at 13:47
  • $\begingroup$ That would certainly suffice. Although I wouldn't translate $x>300$ into $x\ge301$: there's no reason $x$ must be an integer here. $\endgroup$ – Greg Martin Mar 28 '13 at 16:20

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 $z>0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.