Is there any closed form for $\frac{d^{2 n}( \cot z)}{d z^{2 n}}\big|_{z=\frac{\pi}{4}}$? Latest Edit
Thanks to Mr Ali Shadhar who gave a beautiful closed form of the derivative which finish the problem as
$$\boxed{S_n = \frac { \pi ^ { 2 n + 1 } } { 4 ^ { n + 1 } ( 2 n ) ! } | E _ { 2 n } | }, $$
where $E_{2n}$ is an even Euler Number.

In the post, I had found the sum
$$
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{3}}= \frac{\pi^{3}}{32},
$$
and want to investigate it in a more general manner, $$
S_{n}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2 n+1}}
$$
where $n\in N.$
$$
\begin{aligned}
S_{n}&= \lim _{N \rightarrow \infty} \sum_{k=0}^{N} \frac{1}{(4k+1)^{2 n+1}}-\sum_{k=0}^{N} \frac{1}{(4 k+3)^{2 n+1}} \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}-\sum_{k=0}^{N} \frac{1}{\left(k+\frac{3}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}+\sum_{k=0}^{N} \frac{1}{\left(-k-\frac{3}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}+\sum_{k=-N}^{-1} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}}\left[\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}\right] 
\end{aligned}
$$
Using the Theorem:
$$(*):\pi \cot (\pi z)=\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{k+z} ,\quad \forall z \not \in Z.$$
Differentiating (*) w.r.t. $z$ by $2 n$ times yields
$$
\begin{aligned}
& \lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{(-1)^{2 n}(2 n) !}{(k+z)^{2 n+1}}=\frac{d^{2 n}}{d z^{2 n}}[\pi \cot (\pi z)] \\
\Rightarrow & \lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{(k+z)^{2 n+1}}=\frac{\pi}{(2 n) !} \frac{d^{2 n}}{d z^{2 n}}[\cot (\pi z)] 
\end{aligned}
$$
Now we can conclude that
$$\boxed{\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2n+1}}=\left.\frac{\pi^{2n+1}}{4^{2 n+1}(2 n) !} \frac{d^{2 n}}{d z^{2 n}}[\cot z]\right|_{z=\frac{\pi}{4}}}$$
My Question:
Is there any closed form for $\displaystyle \left.\frac{d^{2 n} (\cot z)}{d z^{2 n}}\right|_{z=\frac{\pi}{4}}$?
 A: Using $\csc(z)=\cot(z/2)-\cot(z)$, we have
$$\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(z)=\underbrace{\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\cot(z/2)}_{z=2x\to\, dz=2 dx}-\underbrace{\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\cot(z)}_{0}$$
$$=2^{-2n}\lim_{x\to \frac{\pi}{4}}\frac{d^{2n}}{dx^{2n}}\cot(x)$$
$$\Longrightarrow \lim_{x\to \frac{\pi}{4}}\frac{d^{2n}}{dx^{2n}}\cot(x)=2^{2n}\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(z).$$
Substitute  $\,\,\,\,\displaystyle\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(z)=|E_{2n}|$, (check the edit section at the end)
$$\boxed{\lim_{x\to \frac{\pi}{4}}\frac{d^{2n}}{dx^{2n}}\cot(x)=2^{2n}|E_{2n}|}$$
which matches @Mariusz Iwaniuk's answer in the comments since $(-1)^n E_{2n}=|E_{2n}$|.
A: $$\begin{align}\cot(x+\pi/4)&=\sec2x-\tan2x\\&=\sum_{n\ge0}\frac{(-1)^n}{(2n)!}E_{2n}(2x)^{2n}+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}\cdot2^{2n}(2^{2n}-1)B_{2n}(2x)^{2n-1}\\&=1+\sum_{n\ge1}\frac{(-1)^n2^{2n}}{(2n)!}E_{2n}x^{2n}+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}2^{4n-1}(2^{2n}-1)B_{2n}x^{2n-1}\\\implies\cot z&=1+\sum_{n\ge1}\frac{(-1)^n2^{2n}}{(2n)!}E_{2n}(z-\pi/4)^{2n}\\&+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}2^{4n-1}(2^{2n}-1)B_{2n}(z-\pi/4)^{2n-1}\end{align}$$
You may accordingly extract the even and odd derivatives.
A: This does not answer the question asked in title.
If the goal is to investigate
$$S_{n}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2 n+1}}=\frac 1{4^{2n+1}}\Bigg[\sum_{k=0}^{\infty} \frac 1 {\left(k+\frac{1}{4}\right)^{2 n+1} }  -\sum_{k=0}^{\infty}\frac 1 {\left(k+\frac{3}{4}\right)^{2 n+1} }\Bigg]$$ it could be simple to use directly
$$\sum_{k=0}^{\infty} \frac1 {{(k+a)^m}}=\zeta (m,a)$$ where appears Hurwitz zeta function (which is the generalization of Riemann zeta function).
Then
$$S_n=\frac 1{4^{2n+1}} \left(\zeta \left(2 n+1,\frac{1}{4}\right)-\zeta \left(2
   n+1,\frac{3}{4}\right)\right)$$
