Convergence of series $\sum\limits_{n=1}^{\infty} \frac{((n+1)!)^n}{2!\cdot 4!\cdot \ldots \cdot (2n)}$

Question

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.

Answer 1

You have, for each $n\in\Bbb N$,\begin{align}\frac{(n+2)^{n+1}\cdot(n+1)!}{(2n+2)!}&=\frac{(n+2)^n}{(n+3)(n+4)\cdots(2n+2)}\\&<\frac{(n+2)^n}{(n+2)^{n-1}(2n+2)}\\&=\frac{n+2}{2n+2}\end{align}Now, since$$\lim_{n\to\infty}\frac{n+2}{2n+2}=\frac12<1,$$your series converges.

Answer 2

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