Is there a $k$ such that $a_n=\frac{n^k!}{(n^k!!)^2}$ converges?

Question

Lately I have been playing around with the sequence $$a_n(k) := \frac{n^k!}{(n^k!!)^2}.$$

For $k=1$, it does not look much like it converges.

I don't know $k=2$ it converges, but it doesn't really look like it.

1.

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Question 1. Can we prove that for some $k\in\mathbb{N}$, $a_n=\frac{n^k!}{(n^k!!)^2}$ converges? If not can we prove why not? Thanks mercio.

Also, inspired by my last question, I thought it would be interesting to talk about an extension of $k$ to the complex numbers. $$\frac{n^k!}{(n^k!!)^2}\rightarrow \frac{2^{-1-n^k} \pi \Gamma\left(n^k\right)}{\Gamma\left(1+\frac{n^k}{2}\right)^2}$$

Question 2. Where does this extension of $a_n$ to the complex plane converge? What are its zeroes like?

EDIT: By the way, this is a plot of $a_6(k)$ for $-2\leq \Re(k) \leq 2$, $-2\leq \Im(k) \leq 2$:

(Here, the height is $|a_6(k)|$ and the color is $\operatorname{Arg}{a_6(k)}$.)

It seems that the "stripes" squeeze in as $n$ increases but I don't know what happens in the limit, or if there is a simple explanation of what happens.

Answer 1

As Henry suggests, the sequence always converges to $0$ :

Expanding the factorials, you get :

$(2n)!/(2n)!!^2 = \prod_{k=1}^n (2k-1)/(2k) = \prod_{k-1}^n (1-1/(2k))$ $(2n+1)!/(2n+1)!!^2 = \prod_{k=1}^n (2k)/(2k+1) = \prod_{k-1}^n (1-1/(2k+1))$

In both cases, their logarithm diverges to $- \infty$ (because the harmonic series diverges), and thus the sequence $(n!/n!!^2)$ converges to $0$ (and it should be a $O(n^{-1/2})$ while doing so)