Divergence of an Expression Involving the Gamma Function

Question

I want to show that $$ \lim_{Re(t) \to \infty} \frac{\Gamma(1+2t)\Gamma(-t)}{\Gamma(1+t)} (-z)^t $$ diverges if $|z| > \frac14$. I recognized that the expression involving the Gamma function is just the Beta function, $B(2t+1,-t)$, but I don't know much about the Beta function. Are there some properties of the Gamma function that I can use to prove this divergence? Thanks!

Answer 1

The arguments can be simplified using the Gamma duplication identity $$ \Gamma(2s+1) = \frac{\Gamma(s+1)\Gamma(s+1/2)2^{2s}}{\sqrt{\pi}}, $$ and the Gamma reflection identity $$ \Gamma(s+1)\Gamma(-s) = -\frac{\pi}{\sin(\pi s)}. $$ So your expression is equivalent to $$ -\frac{\sqrt{\pi}}{\sin(\pi s)}\frac{\Gamma(s+1/2)}{\Gamma(s+1)}(-4z)^s $$

The first term in the product has a magnitude in the interval $[\sqrt{\pi}\mathrm{sech}(\mathrm{Im}[s]), \sqrt{\pi}|\mathrm{csch}(\mathrm{Im}[s])|]$ (finite for all $\mathrm{Im}[s]\ne 0$) while the second term can be fairly easily shown from Stirling's approximation to scale as $s^{-1/2}$. So the limit diverges if $|z| > 1/4$ or $\mathrm{Im}[s] = 0$.

Answer 2

Note that some restriction has to be put on $s$ for this limit to make sense: for example, if $s\to\infty$ through positive real numbers, then the $\Gamma(-s)$ term has a pole at every integer, making the limit undefined.

As long as one is willing to restrict, say, the argument of $s$ to an interval like $[\varepsilon,\pi-\varepsilon]$, then this problem goes away and one can use Stirling's formula to asymptotically evaluate each term.