In reviewing Ramanujan's proof of Bertrand's postulate, Ramanujan observes that:
$$\ln\Gamma(x) - 2\ln\Gamma(\frac{x+1}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln(\lfloor\frac{x}{2}\rfloor!)$$
I have noticed that under some circumstances, the following lower bound is better.
$$\ln\Gamma(x) - 2\ln\Gamma(\frac{x+1}{2}) \le \ln\Gamma(x+1) - 2\ln\Gamma(\frac{x+2}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln(\lfloor\frac{x}{2}\rfloor!)$$
For example, this is true when $x=34$. I'm assuming that it is not true in all cases or Ramanujan would have used the tighter lower bound in his proof.
How can I show that $\ln\Gamma(x) - 2\ln\Gamma(\frac{x+1}{2})$ is guaranteed to be lower in all cases $x \ge 2$ and identify the conditions when $\ln\Gamma(x+1) - 2\ln\Gamma(\frac{x+2}{2})$ is the better lower bound?
Here's my thinking on this question:
- $\ln\Gamma(x-1) - 2\ln\Gamma(\frac{x}{2})$ is an increasing function (see here)
- There are some known conditions when $\ln\Gamma(x-1) - 2\ln\Gamma(\frac{x}{2}) \le \ln\Gamma(\lfloor{x}\rfloor+1) - 2\ln\Gamma(\lfloor\frac{x}{2}\rfloor+1)$
- My goal with this question is to show the conditions when $\ln\Gamma(x+1) - 2\ln\Gamma(\frac{x+2}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln\Gamma(\lfloor\frac{x}{2}\rfloor!)$ and when it is not.
Thanks very much,
-Larry
You want to prove $f(x)\leq g(x) \leq h(x)$, where: \begin{align} f(x) &= \ln\Gamma(x) - 2\ln\Gamma\left(\frac{x+1}{2}\right),\\ g(x) &= \ln\Gamma(x+1) - 2\ln\Gamma\left(\frac{x+2}{2}\right),\\ h(x) &= \ln\left( \lfloor{x}\rfloor! \right) - 2\ln\left( \left\lfloor\frac{x}{2} \right\rfloor!\right). \end{align}
Firstly, this is only true for $x\in \mathbb{N}$ with $x\geq 2$. For $x=1$ we have: \begin{align} g(1) &= \ln\Gamma(2) - 2\ln\Gamma\left(3/2\right)\\ &= 2\ln(2) - \ln(\pi)\\ &= 0.24\ldots\\ h(1) &= \ln\left( 1 \right) - 2\ln\left( 1 \right)\\ &= 0, \end{align} so g(1) > h(1). For $x\in \mathbb{R}$, $g(x)>h(x)$ infinitely often.
Secondly, note that for even $x$, $g(x)=h(x)$.
Finally, for odd $x\geq 3$, you can prove $h(x)>g(x)$ by comparing term-by term, and noting that $\Gamma(x/2 + 1)$ is monotonic in this regime ($\Gamma(x)$ has a single minimum at $x=1.46\ldots$).
Comparing $f(x)$ and $g(x)$ is a bit simpler and I'll leave it out for now.