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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 $\displaystyle \left.\frac{d^{2 n} (\cot z)}{d z^{2 n}}\right|_{z=\frac{\pi}{4}}$?

Lai
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3 Answers3

3

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 $(-1)^n E_{2n}=|E_{2n}$|.

Ali Shadhar
  • 25,498
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$$\begin{align}\cot(x+\pi/4)&=\sec2x-\tan2x\\&=\sum_{n\ge0}\frac{(-1)^n}{(2n)!}E_{2n}(2x)^{2n}+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}\cdot2^{2n}(2^{2n}-1)B_{2n}(2x)^{2n-1}\\&=1+\sum_{n\ge1}\frac{(-1)^n2^{2n}}{(2n)!}E_{2n}x^{2n}+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}2^{4n-1}(2^{2n}-1)B_{2n}x^{2n-1}\\\implies\cot z&=1+\sum_{n\ge1}\frac{(-1)^n2^{2n}}{(2n)!}E_{2n}(z-\pi/4)^{2n}\\&+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}2^{4n-1}(2^{2n}-1)B_{2n}(z-\pi/4)^{2n-1}\end{align}$$

You may accordingly extract the even and odd derivatives.

J.G.
  • 115,835
FShrike
  • 40,125
2

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 $$S_n=\frac 1{4^{2n+1}} \left(\zeta \left(2 n+1,\frac{1}{4}\right)-\zeta \left(2 n+1,\frac{3}{4}\right)\right)$$

Claude Leibovici
  • 260,315