Q:

Infinite sums involving $\pi$ and rational numbers.

Can we say anything (i.e. are there any theorems) about "numbers" of the form:
$$S=\sum_{k=1}^\infty\pi^{2k}Q_k,$$ where $Q_k\in\mathbb{Q}$ are monotonicaly decreasing ? We assume that $S$ converges. For example, can we say anything about the "type" of value $S$ is, such as rational, irrational, transcendental, or is there anything "interesting" at all we can say about $S$ ?

A:

If the power series
$$ g(z) = \sum_{k=1}^{\infty} Q_k z^k $$
has radius of convergence larger than $\pi^2$, then $S = g(\pi^2)$. That is, without further conditions on $( Q_k )$ we can say nothing about $S$.
Indeed, we prove that for any positive real number $\alpha > 0$,
$$ A = \left\{ \sum_{k=1}^{\infty} \alpha^{k} Q_k : Q_k \in \Bbb{Q} \text{ and } Q_1 \geq Q_2 \geq \cdots \geq 0 \right\} $$
is equal to $[0, \infty)$. Since $0 \in A$ and $A$ is closed under the multiplication by a positive rational number, it is sufficient to prove that $A$ contains an interval.
Fix a positive rational number $M > 1 + \alpha$, and let $x$ by
$$ x = \sum_{k=1}^{\infty} \frac{\alpha^k}{M^k} = \frac{\alpha}{M-\alpha}. $$
We easily find that both $x$ and $\alpha$ are elements of $A$, and that $x < \alpha$. Now we are going to prove that $(x, \alpha) \subset A$. To this end, fix any $y \in (x, \alpha)$. Then we claim that we can find $\lambda_1, \cdots, \lambda_n \in \Bbb{Q} \cap [0, 1]$ such that
$$ \sum_{k=n+1}^{\infty} \frac{\alpha^k}{M^k} < \color{blue}{\left( y - \sum_{k=1}^{n} \frac{\alpha^k}{M^k} (1 + (M-1)\lambda_k) \right)} < M \sum_{k=n+1}^{\infty} \frac{\alpha^k}{M^k}. \tag{1} $$
It is trivial, by our choice of $y$, such that $(1)$ holds for $n = 0$. We want to prove that $(1)$ also holds for $n+1$ instead of $n$. To this end, let $f(\lambda)$ by
$$ f(\lambda) := y - \sum_{k=1}^{n} \frac{\alpha^k}{M^k} (1 + (M-1)\lambda_k) - \sum_{k=n+1}^{\infty} \frac{\alpha^k}{M^k} (1 + (M-1)\lambda). $$
$(1)$ is equivalent to the condition that $f(0) > 0$ and $f(1) < 0$. Thus there exists $\lambda^* \in (0, 1)$ such that $f(\lambda^{*}) = 0$, and with this $\lambda^{*}$ we have
$$ \sum_{k=n+2}^{\infty} \frac{\alpha^k}{M^k} < \left( y - \sum_{k=1}^{n} \frac{\alpha^k}{M^k} (1 + (M-1)\lambda_k) - \frac{\alpha^{n+1}}{M^{n+1}} (1 + (M-1)\lambda^{*}) \right) < M \sum_{k=n+2}^{\infty} \frac{\alpha^k}{M^k}. $$
Thus we can perturb $\lambda^{*}$ to obtain $\lambda_{n+1} \in \Bbb{Q} \cap [0, 1]$ such that $(1)$ holds for $n+1$, proving our claim.
Now $(1)$ shows that
$$ y = \sum_{k=1}^{\infty} \alpha^k Q_k, \quad Q_k = \frac{1+(M-1)\lambda_{k}}{M^k}. $$
Finally, it is easy to confirm that $Q_k$ is monotone decreasing, hence completes the proof.

