$\Gamma(1/2-n+it)$ converges uniformly

Question

Prove that $\Gamma(1/2-n+it)\rightarrow 0$ uniformly as $n\rightarrow\infty$ for $t\in\mathbb{R}$, where $n$ is a positive integer.

I'm not sure which definition of $\Gamma$ would be easiest to work with, perhaps this one:

$$\Gamma(1/2-n+it)=\dfrac{1}{1/2-n+it}\prod_{m=1}^\infty \dfrac{\left(1+\frac1m\right)^{1/2-n+it}}{1+\frac{1/2-n+it}{m}}$$

How can we go from here?

Answer 1

Note that $\Gamma(1/2-n+it)$ is bounded by $\Gamma(1/2-n)$ by the integral definition, and show that that goes to 0 by any means (like relating it to $\Gamma(1/2)$ by the recursion equation).

Edit: The recursion $\Gamma(x+1)=x\Gamma(x)$.

$\left \lvert\Gamma(1/2-n+it)\right \lvert=\left \lvert \frac{\Gamma(1/2+it)}{(1/2-1+it)(1/2-2+it)\cdots(1/2-n+it)}\right \lvert\leq\frac{1}{2(n-1)!}\left \lvert \Gamma(1/2+it)\right \lvert =\frac{1}{2(n-1)!}\left \lvert \int x^{-1/2t+it}e^x dx\right \lvert\leq \frac{1}{2(n-1)!}\int |x^{-1/2+it}|e^x dx=\frac{1}{2(n-1)!}\int x^{-1/2}e^x dx =\frac{\Gamma(1/2)}{2(n-1)!}$.