Convergence of series $\sum\limits_{n=1}^{\infty} \frac{((n+1)!)^n}{2!\cdot 4!\cdot \ldots \cdot (2n)}$ I am trying to show that the following series is absolutely convergent:
$$\sum\limits_{n=1}^{\infty} \frac{((n+1)!)^n}{2!\cdot 4!\cdot \ldots \cdot (2n)}$$
After writing the denominator as $\prod\limits_{k=1}^{n}(2k)!$ I have tried applying the ratio test which led to
$$\frac{((n+2)!)^{n+1}}{\prod\limits_{k=1}^{n+1} (2k)!}\cdot \frac{\prod\limits_{k=1}^{n}(2k)!}{((n+1)!)^n}=\frac{((n+1)!)^{n+1}\cdot (n+2)^{n+1}}{(2(n+1))!\cdot ((n+1)!)^n}=\frac{(n+2)^{n+1}\cdot(n+1)!}{(2n+2)!}$$
Next I wanted to calculate the limit of this as $n\to\infty$ where I got stuck. WolframAlpha tells me, that the limit is $0$ but I don‘t see where to go from here.
Also: any input on a different approach without the ratio test is greatly appreciated.

$$R_n=\frac{(n+2)^{n+1}\,(n+1)!}{(2n+2)!}$$
$$\log(R_n)=(n+1)\log(n+2)+\log((n+1)!)-\log((2n+2)!)$$ Using twice Stirling approximation and continuing with Taylor series
$$\log(R_n)=n (1-2\log (2))+\left(2-\frac{5 \log (2)}{2}\right)+O\left(\frac{1}{n}\right)$$
$$R_n\sim \frac{e^{n+2}}{2^{\frac{4n+5}{2}} } \quad \to \quad 0$$
