Aim
A particular instance of a rational zeries that has as of yet not been evaluated is:
\begin{align} Z:= \sum_{n=1}^{\infty} \frac{\zeta(2n)}{(2n)!}. \label{EM1} \tag{EM1} \end{align}
This sum can be found in equation (141) of this page. Various equivalent expressions can be obtained. For instance, we have:
\begin{align} Z &= \sum_{n=1}^{\infty} \left( \cosh \left( \frac{1}{n} \right) -1 \right) \label{EM2} \tag{EM2} \end{align}
and \begin{align} Z &= \int_{0}^{\infty} \frac{1}{e^{u} - 1} \ \left[ \frac{ I_{1}(2 \sqrt{u}) - J_{1} (2 \sqrt{u} ) }{ 2 \sqrt{u} } \right] \ du. \label{EM3a} \tag{EM3a} \end{align}
\begin{align} &= \int_{0}^{\infty} \frac{u \ + \ _{0} \tilde{F}_{1} (; 2 ; -u) - \ _{0} \tilde{F}_{1} (; 2 ; u) } {2 (1-e^{u} )} \ du \label{EM3b} \tag{EM3b} \end{align}
Here, $I_{1}(\cdot)$ is the modified Bessel function of the first kind, $J_{1}(\cdot)$ is the Bessel function of the first kind, and $_{p} \tilde{F}_{q}$ is the regularized hypergeometric function. Although these expressions give insights in the nature of $Z$, they do not appear to lead to an evaluation directly. I seek to obtain new information about $Z$, with the eventual aim of evaluating it in closed form. To that end, I propose a method to obtain information about this series, which I've called ``Reverse Euler-Maclaurin Summation.''
Reverse Euler-Maclaurin summation
With this method, we aim at finding a function $f$ such that the fourth term of the E-M formula equates the series described in \eqref{EM1}. Recall that the Euler-Maclaurin formula for a function $f(\cdot)$ that is continuously differentiable $p$ times on the interval ${[}m,n {]}$ is:
\begin{align} \sum_{i=m}^{n} f(i) = \int_{m}^{n} f(x) dx + \frac{f(n)+f(m)}{2} + \sum_{k=1}^{\lfloor p/2 \rfloor} \frac{B_{2k}}{(2k)!}\big{(}f^{(2k-1)}(n)-f^{(2k-1)}(m)\big{)} + R_{p} . \label{EM4} \tag{EM4} \end{align}
The fourth term involves the Bernoulli numbers. These can be recast in terms of the even zeta values:
\begin{align} \zeta(2n) = (-1)^{n+1} \frac{ (2 \pi)^{2n} B_{n} }{2 (2n)! } \implies B_{2n} = \frac{2 \zeta(2n) (2n)!}{ (2 \pi)^{2n}} (-1)^{-n-1} \label{EM5} \tag{EM5} \end{align}
When we substitute the latter expression in the fourth term of the Euler-Maclaurin formula, we obtain:
\begin{align} \sum_{k=1}^{\lfloor p/2 \rfloor} \frac{B_{2k}}{(2k)!}\big{(}f^{(2k-1)}(n)-f^{(2k-1)}(m)\big{)} = \sum_{k=1}^{\lfloor p/2 \rfloor} \left( \frac{2 \zeta(2k) }{ (2 \pi )^{2k} } (-1)^{-k-1} \right) \big{(}f^{(2k-1)}(n)-f^{(2k-1)}(m)\big{)}. \label{EM6} \tag{EM6} \end{align}
In order to find a new expression for $Z$ as described in equation \eqref{EM1}, we therefore aim at finding a function $f \in C^{\infty}$ such that it satisfies:
\begin{align} f^{(2k-1)} (1) = \frac{(2 \pi)^{2k} }{2 \ (2k)! } (-1)^{k+1} \ \text{ and } \ f^{(2k-1)} (\infty) = 0. \quad \text{ for all } k \in \mathbb{Z}_{\geq 1} \label{EM7} \tag{EM7} \end{align}
Moreover, we seek to find an $f$ such that $R_{p} \to 0$ as $p \to \infty$.
Delay differential equations
I would like to solve the two systems of differential-like equations in \eqref{EM7}. Although I don't know how to solve them simultaneously, yet, it appears to be somewhat similar to several delay differential equations I've found over here. For instance, in this question, the equation
\begin{align} g^{(n)} (0) = g(n). \label{EM8} \tag{EM8} \end{align}
is studied. If we set $W(\cdot)$ as the principal branch of the Lambert W function and put $c:= -W(-1)$, then user GEdgar obtains the elegant solution:
\begin{align} g(x) = e^{cx} \cos(cx). \label{EM9} \tag{EM9} \end{align}
Moreover, in this question and this one, the equation
\begin{align} \frac{h^{(n)}(0)}{n!} = h(n) \label{EM10} \tag{EM10} \end{align}
is considered. User abacaba obtains a solution in terms of a rational linear combination of $\operatorname{sinc}$-like functions:
\begin{align} h(z) = \sum_{i=0}^{\infty} a_{i} \phi_{i}(z), \label{EM11} \tag{EM11} \end{align}
where \begin{align} \phi_{i}(z) = (-1)^{i} \frac{\sin(\pi z)}{\pi (z-i)} \label{EM12} \tag{EM12} \end{align}
and $a_{i} \in \mathbb{Q} \setminus \{ 0 \}$. Results like these give me hope that the two systems of equations in \eqref{EM7} can simultaneously be solved as well. If and once this solution is obtained, it can be substituted in the Euler-Maclaurin formula \eqref{EM4}. I hope it provides information about $Z$ in \eqref{EM1}.
Power series solution
We guess we can obtain a solution by means of a power series of the form:
\begin{align} f(x) := \sum_{n=1}^{\infty} c_{n}(x-1)^{n}. \label{EM13} \tag{EM13} \end{align}
Given the first of the two systems of equations in \eqref{EM7}, we find
\begin{align} c_{2n-1} = \left( \frac{1}{2} \right) \frac{ (2 \pi)^{2n} }{ (2n-1)! \ (2n)! } (-1)^{n+1} \ \text{ and } \ c_{2n} = 0. \label{EM14} \tag{EM14} \end{align}
Therefore, we have
\begin{align} f(x) = - \frac{ \sqrt{ \pi } \left( \operatorname{bei}_{1} (2 \sqrt{2 \pi (x-1)}) + \operatorname{ber}_{1} ( 2 \sqrt{ 2 \pi (x-1) } ) \right) }{ 2 \sqrt{x-1}} \label{EM15} \tag{EM15} \end{align}
Here, $\operatorname{bei}_{1}(\cdot)$ and $\operatorname{ber}_{1}(\cdot)$ are the first-order Kelvin bei and ber functions, respectively.
Convergence issues
It seems, however, that substitution of this function in the sum and integral terms of the Euler-Maclaurin summation equation -- see \eqref{EM4} -- do not converge with this choice of $f(\cdot)$. I suspect this is the case because the condition $f^{(2k-1)} (\infty) = 0$ is not satisfied. In fact, it appears to be the case that for the $f$ we have constructed, we have $f^{(2k-1)} (\infty) = \pm \infty$ for all $k \in \mathbb{Z}_{\geq 1}$.
Questions
- Can the $f(\cdot)$ obtained in equations \eqref{EM13} -- \eqref{EM15} be modified, so it also satisfies the second system of equations in \eqref{EM7}?
- Can one obtain other solutions that are continuously differentiable on $[1, \infty] $ and satisfy the two systems of equations in \eqref{EM7} simultaneously? If so, how does one construct them?
