On the wiki on rational zeta series, it is stated that - given a real number $x$ - it can be represented as follows:
$$ x = \sum_{n=2}^{\infty} q_{n} \zeta(n,m). \tag{1} $$ Here, $q_{n}$ is a sequence of rational numbers, the value $m$ is fixed, and $\zeta(s,m)$ is the Hurwitz zeta function. It is furthermore asserted that "it is not hard to show that any real number $x$ can be expanded in this way".
Evidently being a less talented mathematician than the author of the piece, I haven't been able to find a proof of this assertion yet. I did find a way to expand any rational number $y$ as described in equation $(1)$. We can do so as follows:
Observe that $$\sum_{n=2}^{\infty} \zeta(n,2) = \sum_{n=2}^{\infty} [\zeta(n)-1] =1. \tag{2} $$
If we then proceed to multiply both sides of the equation with a rational number $y=a/b \in \mathbb{Q} $, we obtain \begin{align*} \frac{a}{b} &= \frac{a}{b} \sum_{n=2}^{\infty} [\zeta(n)-1] \newline &= \sum_{n=2}^{\infty} \Big{(} \frac{a}{b} \Big{)} [\zeta(n)-1] . \tag{3} \end{align*}
So now we have $q_{n} = a/b$ for all $n \geq 2$.
Questions
- How to prove that any number $x \in \mathbb{R}$ can be represented as described in equation $(1)$ ?
- Suppose we restrict ourselves to the case $m=2$. Is it then still possible to expand every real number this way?
