Prove absolute convergence of $\sum_{n=1}^{\infty} \frac{2^nn!}{(z+1)(z+3)...(z+2n+1)}$

Question

I need to prove that

$$\sum_{n=1}^{\infty} \frac{2^nn!}{(z+1)(z+3)...(z+2n+1)}, \operatorname{Re}z \gt \frac{1}{2}$$

is absolutely convergent.

I've tried to use d'Alambert ratio test, but I got $$\lim\limits_{n\to \infty}\vert{\frac{a_{n+1}}{a_n}}\vert = \lim\limits_{n\to \infty}\vert\frac{2^{n+1}(n+1)!(z+1)(z+3)...(z+2n+1)}{2^nn!(z+1)(z+3)...(z+2n+1)(z+2n+3)}\vert= 1$$ and so it's inconclusive.

Also I've tried to rewrite denominator as $\frac{(z+2n+1)!!}{(z-1)!!}$. Then I think I can use the fact that $(2k+1)!! = \frac{(2k+1)!}{2^kk!}$, the property of gamma function that $Γ(z+1)=zΓ(z)$ for $\forall z \in \mathbb{C}$ and the Stirling formula for gamma function, but I don't know how to apply all of this correctly there. How should I do it?

Or can you give any other ideas how to prove this?

Thanks for your help.

Answer 1

For any $z$ such that $\text{Re}(z)>1$ the given series equals

$$ \frac{1}{z+1}\sum_{n\geq 1}\frac{n!}{\left(\frac{z+1}{2}+1\right)\cdots\left(\frac{z+1}{2}+n\right)}=\frac{1}{2}\sum_{n\geq 1}\color{blue}{\frac{\Gamma(n+1)\,\Gamma\left(\tfrac{z+1}{2}\right)}{\Gamma\left(\frac{z+3}{2}+n\right)}}$$ or $$ \frac{1}{2}\sum_{n\geq 1} B\left(n+1,\frac{z+1}{2}\right) = \frac{1}{2}\int_{0}^{1}(1-u)^{\frac{z-1}{2}}\sum_{n\geq 1}u^n\,du $$ or $$ \frac{1}{2}\int_{0}^{1}u (1-u)^{\frac{z-3}{2}}\,du=\frac{1}{2}B\left(2,\frac{z-1}{2}\right)=\color{red}{\frac{2}{z^2-1}} $$ and the absolute convergence follows by applying Gautschi's inequality to the blue term.
On the other hand we cannot have absolute convergence for any $z$ with real part in $\left(\frac{1}{2},1\right]$ because $\frac{2}{z^2-1}$ has a pole at $z=1$!