Question about Ramanujan's proof of Bertrand's Postulate I am reviewing Ramanujan's proof of Bertrand's Postulate which can be found here.
At step #7, he writes: "But it is easy to see that..."
$\log\Gamma(x) - 2\log\Gamma\left(\frac{1}{2}x + \frac{1}{2}\right) \le \log[x]! - 2\log\left[\frac{1}{2}x\right]!$
As $\Gamma(x+1) = [x]!$, this seems reasonable enough to me.
My question is: why did Ramanujan choose $\frac{1}{2}$.  It seems like this would be true for any $n \ge 0$ where:
$\log\Gamma(x) - 2\log\Gamma\left(\frac{1}{2}x + n\right) \le \log[x]! - 2\log\left[\frac{1}{2}x\right]!$
Is that correct? 
Thanks,
-Larry
 A: Inequality (7) is delicate. $\log x!-\log\Gamma(x)$ is a little bigger than $2\log (x/2)!-2\log\Gamma(x/2+1/2),$ which makes the lower bound tight. $\log\Gamma(x+1)$ is no different from $\log x!$ so he relies on $2\log\Gamma(x/2+1/2)$ being slightly less than $2\log(x/2)!$ to achieve the upper bound.
In (8) he refers to the Stirling approximation for the bounds in (7), so there may have been an element of computational convenience in the choice 1/2. Once he establishes the numerical bound in (8),(9) he has no further use for the $\log\Gamma$ functions used in (7).
There may be room for a constant other than 1/2 but anything you do will have an effect at both ends. $2\log\Gamma(x/2+3/4)$ might be closer to and still less than $2\log(x/2)!$ but will the left inequality hold? It appears not.
Whether 1/2 is optimal I am not sure but I would guess so. When you examine the asymptotic behavior of $\log\Gamma$ the derivative becomes important. Question 94582, though I can't link to it at the moment, concerned this property. The factor 1/2 in argument of $\log\Gamma$ (and 2 outside it) make it something to check as an explanation of the choice of 1/2.
Edit: In light of this question it appears that $1/2$ is needed to give desired asymptotic behavior to the l.h.s. and r.h.s. For completeness:
$PG(x) :=$ Polygamma(x). From (7):
$$\ln\Gamma(x) -2\ln\Gamma\left(\frac{x+1}{2}\right)\leq \cdot \leq \ln\Gamma\left(x+1\right)-2\ln\Gamma\left(\frac{x+1}{2}\right) $$
$$\frac{d}{dx}\left(\ln\Gamma(x)-2\ln\Gamma\left(\frac{x+1}{2}\right)\right)= PG(x)-PG\left(\frac{x+1}{2}\right)\sim \ln 2^-$$
$$\frac{d}{dx}\left(\ln\Gamma\left(x+1\right)-2\ln\Gamma\left(\frac{x+1}{2}\right)\right)= PG\left(x+1\right)-PG\left(\frac{x+1}{2}\right)\sim \ln 2^+$$
So $\log[x]! -2\log[x/2]!$ is trapped between two functions which approach $\log 2$ from above and below. The 1/2 is from $\frac{x+1}{2}$ and is essential to the asymptotic limits.
