On the sets of sums $\sum\limits_{n=1}^\infty\frac{a_n}{n^s}$ with $(a_n)$ periodic and integer valued, for different values of $s$ natural number

Question

For every positive integer $s$, let $A_s$ denote the set of the sums of the converging series $\sum\limits_{n=1}^\infty\frac{a_n}{n^s}$ for every periodic sequence of integers $(a_n)$.

Then each $A_s$ is a countable dense subset of the real numbers, and an additive group. The set $A_1$ is in fact a vector space with scalars drawn from the rationals.

I suspect $A_s$ should contain no non-zero rationals (counterexamples are welcome!) but a proof of this would imply that Catalan's number is irrational so attacking that directly should be avoided...

Question Can anything interesting be said about the intersections of these sets? For example, is it the case that $A_s\cap A_t=\{0\}$ for every $s\ne t$?

This question comes from my own musings and it may be open. I suppose this is a risk one always has when asking questions that flirt with the zeta function.

Some Notes: $\zeta(s)\in A_s$, $\eta(s) \in A_s$, $\ln(\mathbb{Q})\subset A_1$.

Generalizations that may be worthy of follow up:

1) Is this just the case for positive real numbers $s\neq t$? This has now been answered below. This is not the case.

2) If we define $A_s$ with Gaussian integers do we get the same results?

Edit 1 (an effort to spruce this question up): Some Motivations + some cool values

This question didn't get the excitement I expected so I will now add some crazy values! Here are a couple of values from Dirichlet series in $A_1$, $A_3$, $A_5$, $A_7$. We can compute specific values in $A_s$ but when we manage to get exact forms of values in these sets (it seems) invariably this is because of their relationship to Dirichlet Series.

$$f(s,\vec{a})= \sum_{n=1}^\infty{\frac{a_n}{n^s}} $$

Then $$ \begin{array}{c|c|c|c|c|c} f(s,\vec{a}) & \vec{a}=(1,-1) & \vec{a}=(1,0,-1,0) & \vec{a}=(1,1,0,-1,-1,0) & \vec{a}=(1,0,1,0,-1,0,-1,0) \\ \hline %%%%%%%%%%%%%%%%%%%%%%%%%%%%% s=1 & \ln(2) & \frac{\pi}{4} & \frac{2 \pi}{3\sqrt{3}} & \frac{\pi}{2\sqrt{2}} \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%% s=3 & \frac{3}{4}\zeta(3) & \frac{\pi^3}{32} & \frac{5 \pi^3}{81\sqrt{3}} & \frac{3\pi^3}{64\sqrt{2}} \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%% s=5 & \frac{15}{16}\zeta(5) & \frac{5 \pi^5}{1536} & \frac{17 \pi^5}{2916\sqrt{3}} & \frac{19 \pi^5}{4096 \sqrt{2}} \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%% s=7 & \frac{63}{64}\zeta(7) & \frac{61\pi^7}{184320} & \frac{91 \pi^7}{157464\sqrt{3}} & \frac{307 \pi^7}{655360\sqrt{2}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{array}$$

Column(1) Column(2) Column(3) Column(4) And more

So here are just some specific elements in $A_s$ to get a feeling for these sets.

Answer 1

I sought out an authority of some kind on these matters and I will reproduce the answer below. To summarize, the answer is that my conjecture is the expectation of mathematical community but there doesn't seem to be much in the way of evidence for these expectations.

This comes from an email exchange with Professor Wadim Zudilin who studies these type of mathematical structures. I have added some formatting but really have left the content untouched.

Dear Mason,

I cannot be long in my answer but indeed there are some expectations on how the sets $A_s$ in your notation are structured for positive integers $s$. I assume that $\{a_n\}$ is periodic from the very beginning: $a_k=a_{T+k}$ for all $k=1,2,\dots,$ with $T$ a fixed period. The only $\mathbb{Q}$-linear relations within $A_s$ for $s$ fixed are expected to be those evaluating to $0$ (that is, no rational numbers apart from $0$ can be in the $\mathbb{Q}$-linear span of $A_s$). Furthermore, $A_s$ and $A_t$ are linearly disjoint for $s\ne t$ in the sense that their $\mathbb{Q}$-linear spans intersect at $\{0\}$. A usual language of dealing with $A_s$ is through the Hurwitz zeta function, and there very few results to support those great expectations. There was some work on $A_1$ in relation with Baker's linear forms in logarithms; the latter methods imply that if a $\mathbb{Q}$-linear combination of the elements in $A_1$ is irrational then it is transcendental as well. This is not quite what your question is about, except that we would believe that being irrational means being nonzero.

That's all I can tell you. Don't blame professional mathematicians for not proving somewhat definite towards your (quite natural) expectations: it is extremely hard proving that the numbers are linearly independent over the rationals, because one needs to create for that very good rational approximations to those numbers. And the series in $A_s$ converge too slowly to be easily approximated by the rationals...

Best wishes, Wadim Zudilin

Answer 2

This is a response to the note from user90369. What is the protocol for this?

Thanks for your comment.

What was written originally wasn't quite my conjecture but the current rewrite seems to match my conjecture. Note that $\vec{a}=(1,-1,-2,-1,1,2), s=1, \vec{c}=\vec{0}, t\in \mathbb{R^+}$ would satisfy $f(s,\vec{a})=f(t,\vec{c})$ which can be seen here.

Also the wording no solution has made me realize that this seems to be unlikely. There will often be some solution in the real numbers. My conjecture is that there shouldn't be a solution for $s$ in the positive integers.

We might be able to defeat the broader question about what is happening in the real numbers by arguing that $f(t,\vec{c})$ is a continuous map in $t$ onto it's range and $f(t,\vec{a})$ is somewhere in this range. That seems like it should be a fruitful approach but it doesn't really help answer much about $t\in\mathbb{N}$. It might be the case that these zeta-like functions don't have any particularly special relationship with $s\in \mathbb{N}$ so that would make my question tricky to answer (That seems unlikely to me). But that would be kind of an interesting answer though...

Here is a concrete example. Take $\vec{a}=(1 ,-3,1,1)$ and $\vec{c} =(1,0,-1,0)$. Then $f(s,\vec{c})$ is the dirichlet beta function. $f(1,\vec{a})=0$ and I have written a tiny bit on this case here.

Then $$\frac{\pi^3}{32}=f(3,\vec{c})=f(s^*,\vec{a})$$

where $s^*\approx 6.554$ for a picture of what is happening here check out this. For more digits you can check out this.

Without being able solve for the exact value of $s^*$ we know that $0<\frac{\pi^3}{32}<1$ is in the range of $f(t,\vec{a})$ because $f(1,\vec{a})=0$ and $\lim_{t\to \infty}f(t,\vec{a})=1$. So if $f(t,\vec{a})$ is a continuous function in $t$ then it must take on the value of $\pi^3/32$ somewhere.