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!



Closed, sum-free form for the $n$-th derivative of $\operatorname{arcsinh}(\frac1x)$ in $x=1$
Here is a proof that $M\_{nn}=O(n^{-1})$, as the OP conjectured. The function
$$f(z):=\operatorname{arcsinh}(1/z),\qquad |z|<1,$$
is holomorphic in the unit disk, hence by Cauchy's formula
$$ \frac{f^{(k)}(1)}{k!}=\frac{1}{2\pi i}\int\_{|z-1|=r}\frac{f(z)}{(z-1)^{k+1}}\,dz,\qquad 0<r<1.$$
Here and later circles are positively oriented. It follows, by applying the binomial formula, that
$$\frac{1}{2}M\_{nn}=\frac{1}{2\pi i}\int\_{|z-1|=r}\frac{f(z)}{z-1}\left(\frac{z+1}{z-1}\right)^n\,dz,\qquad 0<r<1.$$
We make the change of variables $z:=(1+w)/(1-w)$, keep track of the image of the original circle (which is another circle), and deform it to a circle centered at the origin. This way we obtain
$$\frac{1}{2}M\_{nn}=\frac{1}{2\pi i}\int\_{|w|=\rho}\frac{g(w)}{1-w}\cdot\frac{dw}{w^{n+1}},\qquad 0<\rho<1,$$
where $g(w)$ is the following holomorphic function in the unit disk:
$$g(w):=\operatorname{arcsinh}\left(\frac{1-w}{1+w}\right),\qquad |w|<1.$$
This way we see, again by Cauchy's formula, that $M\_{nn}$ is twice the sum of the first $n$ Taylor coefficients of $g(w)$ at the origin. (**Added:** This can also be derived without complex analysis, as in Max Alekseyev's response.) Using the identities
$$g'(w)=\frac{-\sqrt{2}}{(1+w)(1+w^2)^{1/2}},\qquad|w|<1,$$
$$\frac{1}{(1+w^2)^{1/2}}=\sum\_{k=0}^\infty\frac{(-1)^k}{4^k}\binom{2k}{k}w^{2k},\qquad|w|<1,$$
it is straightforward to derive that
$$\frac{1}{2}M\_{nn}=\operatorname{arcsinh}(1)-\sqrt{2}\sum\_{2k\leq n}\frac{(-1)^k}{4^k}\binom{2k}{k}\sum\_{2k\leq m\leq n-1}\frac{(-1)^m}{m+1}.$$
The inner sum is positive and it is of size $O(k^{-1})$, hence in the outer sum the $k$-th term has sign $(-1)^k$ and it is of size $O(k^{-3/2})$. It follows that $M\_{nn}$ converges. By applying a more careful asymptotic analysis in $k$ and keeping track of the error term $O(n^{-1})$ coming from the inner sum, the convergence is seen to occur with speed $O(n^{-1})$. Finally, by [Abel's theorem](https://en.wikipedia.org/wiki/Abel%27s_theorem), 
$$\lim\_{n\to\infty} M\_{nn}=2g(1)=0,\qquad\text{hence in fact}\qquad M\_{nn}=O(n^{-1}).$$


