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My question relates to this step in the proof here:

But it is easy to see that

$$\log \Gamma(x)-2\log\Gamma(\frac12x+\frac12) \le \log\left\lfloor x\right\rfloor!-2\log\left\lfloor\frac12x\right\rfloor!$$

if I use $\left\{\dfrac{x}{2}\right\} = \dfrac{x}{2} - \left\lfloor\dfrac{x}{2}\right\rfloor$ to represent the fractional part.

It is clear to me that the statement is true for when $\left\{\dfrac{x}{2}\right\} \le \dfrac{1}{2}$. I am having trouble understanding why it is necessarily true when $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$

When $\left\{\dfrac{x}{2}\right\} \le \dfrac{1}{2}$, I can use the answer from here.

where

$x_1 = x$, $\Delta{t_1} = 1-\left\{x\right\}$, $x_1+\Delta t_1 = \lfloor{x}\rfloor+1$

$x_2 = \dfrac{x}{2}+\dfrac{1}{2}$, $\Delta{t_2} = \dfrac{1}{2}-\left\{\dfrac{x}{2}\right\}$, $x_2+\Delta t_2 = \lfloor\dfrac{x}{2}\rfloor+1$

$x_3 = \dfrac{x}{2}+\dfrac{1}{2}$, $\Delta{t_3} = \dfrac{1}{2}-\left\{\dfrac{x}{2}\right\}$, $x_3+\Delta t_3 = \lfloor\dfrac{x}{2}\rfloor+1$

to get:

$$\dfrac{\Gamma(\left\lfloor{x}\right\rfloor+1)}{\Gamma({x})} \ge \frac{\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1)}{\Gamma(\dfrac{x}{2}+\dfrac{1}{2})}\dfrac{\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1)}{\Gamma(\dfrac{x}{2}+\dfrac{1}{2})}$$

This approach fails for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$ since $\Delta{t_2}, \Delta{t_3} < 0$

Can someone provide me the argument for why this inequality is true for the condition where $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$?

Thanks very much.


Edit: I figured out an argument that works for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$

I posted it as the answer below.

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  • $\begingroup$ Hi Daniel, I believe that when Ramanujan used $\psi$, it was the Second Chebyshev Function and not the Digamma Function. If you can explain more how these functions relate, that would help me to understand your point. $\endgroup$ – Larry Freeman Jan 25 '15 at 22:34
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    $\begingroup$ Yes unfortunate about the notation. It is not the digamma function. Ramanujan's version of this proof is clearer IMO (I looked at the one you are working from). The main definition that is omitted from your version is $\log[x]! = \psi(x)+\psi(x/2)+\psi(x/3)+...$ I have to run but glad to help if someone else doesn't beat me to it. $\endgroup$ – daniel Jan 26 '15 at 3:54
  • $\begingroup$ Thanks for clarifying, I'll think about. That might be the point that I was missing. $\endgroup$ – Larry Freeman Jan 26 '15 at 4:27
  • $\begingroup$ I think I got it. $\endgroup$ – Larry Freeman Jan 26 '15 at 16:23
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I figured out an argument that works for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$

$$\log\Gamma(x) \le \log\Gamma(x + 1 - \left\{ x\right\}) = \log\Gamma(\left\lfloor x\right\rfloor + 1) = \log\left\lfloor x\right\rfloor!$$

$$2\log\Gamma(\dfrac{x}{2} + \dfrac{1}{2}) \ge 2\log\Gamma(\dfrac{x}{2}+1 - \left\{\dfrac{x}{2}\right\}) = 2\log\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1) = 2\log\left\lfloor\dfrac{x}{2}\right\rfloor! $$

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