During research involving the Born–Jordan quantization I came across the expression

$$ \frac{d^k}{dx^k}\operatorname{arcsinh}\Big(\frac1x\Big)\Big|_{x=1}\tag1 $$

for $k\in\mathbb N_0$. It is not too hard to write this expression as a sum

$$ (1)=\frac{\sqrt{2}}{2^k}\sum_{j=0}^{k-1} a_j^k\tag{2a} $$

for any $k\in\mathbb N$ where $(a_j^k)_{j\in\mathbb Z,k\in\mathbb N}$ is a recursive sequence of integers given by

$$ a_j^k:=\begin{cases} a_0^1=-1&\\a_j^k=0&\text{if }j<0\text{ or }j\geq k\\ a_j^{k+1}=a_j^k(2j-k)+a_{j-1}^k(2j-3k-1)&\text{else}\end{cases},\tag{2b} $$

(basically a modified version of Pascal's triangle). Unfortunately, I so far was not able to find a closed *(sum-free)* form of $(1)$ / $(2a)$ for arbitrary $k\in\mathbb N_0$.

This recursive sequence is nice for explicit calculations (especially speeds up things for larger $n$) - but I'm interested how $(1)$, or rather of the arising matrix elements

$$ M_{nn}:=2\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{2^k}{k!}\Big(\frac{d^k}{dx^k}\operatorname{arcsinh}\Big(\frac1x\Big)\Big|_{x=1}\Big)\tag3 $$

behave for large $n$.

Explicit calculations (up until $n=200$) suggest that $M_{nn}\overset{n\to\infty}\longrightarrow0$ with $M_{nn}=\mathcal O(\frac1n)$ for $n\to\infty$. What could be an approach to potentially prove this? What is more realistic: trying to find some bound for $(3)$ or trying to find a closed, sum-free form for $(1)$, respectively $(2a)$ / $(2b)$?

Being fairly new here I hope this "question" (or rather problem) is suitable for mathoverflow. If it is not, feel free to tell me so I can outsource it to math.stackexchange. Thanks in advance for any answer or comment!

(I also changed my title to clarify the "sum-free" part here which before "only" was part of the full question)$\endgroup$ – Frederik vom Ende Mar 12 '18 at 17:18