There is a known proof of the identity $$\sum_{k=0}^\infty \frac1{(2k+1)^2}=\frac{\pi^2}8\,\,\,(*)$$ (equivalent to $\sum \frac1{n^2}=\frac{\pi^2}6$) by expanding $\arcsin x$ as a power series $$\arcsin x=\sum_{k=0}^\infty \frac{(2k-1)!!}{(2k)!!(2k+1)}x^{2k+1},$$ substituting $x=\sin t$ and integrating for $t$ from 0 to $\pi/2$. The double factorials cancel by some magic and we get $(*)$.

I wonder whether the higher values of zeta at even integer points may be obtained on this way for other functions similar to $\arcsin$ or other clever integrals.

  • 2
    $\begingroup$ One of the standard proofs of the formulas for $\zeta(k)$ uses power series of $\tan$. What is special about $\arcsin$, why do you insist on using it? $\endgroup$ – Alexandre Eremenko Dec 6 '17 at 20:18
  • $\begingroup$ it uses not only power series, but a sum of fractions, right? $\endgroup$ – Fedor Petrov Dec 6 '17 at 20:50

In $[1]$ Borwein and Chamberland show that for $|x|\le 2$ \begin{align*} \arcsin^{2N+1}(x/2) &= (2N+1)!\sum_{k=0}^\infty \frac{G_N(k){2k\choose k}}{2(2k+1)4^{2k}} x^{2k+1}, & N = 0,1,\ldots \tag{1} \\ \arcsin^{2N}(x/2) &= (2N)!\sum_{k=1}^\infty \frac{H_N(k)}{{2k\choose k}k^2} x^{2k} & N = 1,2,\ldots \tag{2} \end{align*} where \begin{align*} G_0(k) &= 1 \\ G_N(k) &= \sum_{n_1=0}^{k-1}\frac{1}{(2n_1+1)^2} \sum_{n_2=0}^{n_1-1}\frac{1}{(2n_2+1)^2} \cdots \sum_{n_{N}=0}^{n_{N-1}-1}\frac{1}{(2n_{N}+1)^2} \end{align*} and \begin{align*} H_1(k) &= 1/4 \\ H_N(k) &= \frac{1}{4} \sum_{n_1=1}^{k-1}\frac{1}{(2n_1)^2} \sum_{n_2=1}^{n_1-1}\frac{1}{(2n_2)^2} \cdots \sum_{n_{N-1}=1}^{n_{N-2}-1}\frac{1}{(2n_{N-1})^2}. \end{align*} Letting $x=2\sin t$ in $(1)$ and $(2)$ and integrating from $0$ to $\pi/2$ we find the same type of cancellation described in the question statement, with the result \begin{align} \sum_{k=0}^\infty \frac{G_N(k)}{(2k+1)^2} &= \frac{1}{(2N+2)!}\left(\frac{\pi}{2}\right)^{2N+2} \tag{3} \\ \sum_{k=1}^\infty \frac{H_N(k)}{k^2} &= \frac{1}{(2N+1)!}\left(\frac{\pi}{2}\right)^{2N}. \tag{4} \end{align} Equations $(3)$ and $(4)$ are natural generalizations of $(*)$. They give the value of $\zeta(2)$ and relate the values of $\zeta(n)$ for $n=4,6,\ldots$ to other interesting sums.

Equation $(3)$ implies for $N=0,1$ that \begin{align*} \sum_{k=0}^\infty \frac{1}{(2k+1)^2} &= \frac{\pi^2}{8} \\ \sum_{k=0}^\infty \frac{\psi^{(1)}(k+1/2)}{(2k+1)^2} &= \frac{5\pi^4}{96}, \end{align*} where $\psi^{(n)}(k)$ is the polygamma function. The last result can be reexpressed as $$\sum_{k=1}^\infty \frac{1}{k^4} = \frac{\pi^4}{72} - \frac{4}{15}\sum_{k=0}^\infty\frac{\psi^{(1)}(k+3/2)}{(2k+1)^2}.$$ (Given $\zeta(4) = \pi^4/90$ this implies $\sum_{k=0}^\infty{\psi^{(1)}(k+3/2)}/{(2k+1)^2} = {\pi^4}/{96}.$)

Equation $(4)$ implies for $N=1,2$ that \begin{align*} \sum_{k=1}^\infty \frac{1}{k^2} &= \frac{\pi^2}{6} \\ \sum_{k=1}^\infty \frac{H_{k-1}^{(2)}}{k^2} &= \frac{\pi^4}{120}, \end{align*} where $H_k^{(n)}$ are the generalized harmonic numbers of order $n$. The last result can be reexpressed as $$\sum_{k=1}^\infty \frac{1}{k^4} = \sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2} - \frac{\pi^4}{120}.$$ (Given $\zeta(4)$ this implies $\sum_{k=1}^\infty{H_k^{(2)}}/{k^2} = {7\pi^4}/{360}.$)

$[1]$ Borwein, J. M. and Chamberland, M. "Integer Powers of Arcsin." Int. J. Math. Math. Sci., Art. 19381, 1-10, 2007.

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    $\begingroup$ oh, that is exactly what I wanted to see! $\endgroup$ – Fedor Petrov Jul 1 '18 at 20:29

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