# Is this class of series all demonstrably transcendental?

Question:

For a vector with integer entries $[a_0, a_1, \dots, a_{k-1}]$ is it true that when $\sum_{n=1}^\infty{\frac{a_{n-1 \mod k}}{n}}$ is not divergent it limits to some transcendental number or zero?

Musings:

I will adopt something like the notation of this post. We might call these Weinberger series. Err... I dunno we might call them something else. Let $\vec{v}=[a_0, a_1, \dots a_k]$ be a vector with integer entries.

$\sum{\vec{v}}=[\overline{a_0,a_1, \dots, a_{k-1}}]=\sum_{n=1}^\infty{\frac{a_{n-1 \mod k}}{n}}$. I should say that I suspect that when the sum of entries of $\vec{v}$ is not zero we have that $\sum{\vec{v}}$ is divergent. All the following have that property that the sum of entries is zero (This makes the 4th entry not ambiguous).

Let me show you a few! In this notation:

$\begin{array}{lclr} \\ \frac{\pi\sqrt{2}}{4} & = & [\overline{1,0,1,0,-1,0,-1,0}] & \text{Why [1]} \\ \frac{\pi\sqrt{3}}{9} & = & [\overline{1,-1,0}] & \text{Don't [2]} \\ \frac{\pi\sqrt{7}}{7} & = & [\overline{1,-1,-1,1,-1,1,0}] & \text{Hyperlinks [3]} \\ \ln{k} & = & [\overline{1,1,\dots,1, 1-k}] & \text{Work[4]} \\ \frac{\sqrt{3}\pi+3\ln\left(2\right)}{9} & = & [\overline{1,0,0,-1,0,0}] & \text{In [5]} \\ \frac{\pi+2\coth^{-1}\left(\sqrt{2}\right)}{4\sqrt{2}} & = & [\overline{1,0,0,0,-1,0,0,0}] & \text{Arrays [6]} \end{array}$

Why 1 don't 2 hyperlinks 3 work 4 in 5 arrays 6?

I suspect that these are all transcendental when they are not $0$ or $\infty$. In fact! I am hoping to be able to say that they all fit neatly into some class. They all look to be $\alpha \pi+ \beta\ln(\gamma)+\delta$ for some algebraic constants $\alpha, \beta, \gamma, \delta$. But I would settle for just seeing that the guys need to be transcendental (or some clever counterexample that I am missing.) I suspect that their periodic nature should give rise to a demonstration that these are not algebraic numbers.

How can I do that?

Let me defend my use of $\vec{v}$. One should only use this notation if they are vectors is some sense. And they are. Note that we can define a type of scalar multiplication with the rationals so that

$$\frac{3}{5}\ln(2)= \frac{3}{5} [\overline{1, -1}] = [\overline{0,0,0,0,3,0,0,0,0,-3}]$$

This is really not me saying much more than

$$\frac{3}{5}\sum_{n=1}^\infty{\frac{(-1)^{n+1}}{n}}=\sum_{n=1}^\infty\frac{3(-1)^{n+1}}{5n}$$

We have all the properties that one desires of a vector space: These values are closed under addition and have a type of multiplication with rational numbers. It leaves me wondering what the right type basis should be for this type of exploration.

• I haven't worked through the proof but theorem 1 of this implies the claim. Commented Jul 5, 2022 at 22:14

Your claim seems to be essentially equivalent to the statement that the numbers $\Psi(i/k)+\gamma$ for $i=1\ldots k-1$ and their nonzero linear combinations over the rationals are transcendental. I suspect this is true, but I don't know if it can be proven.

EDIT:
Gauss's digamma theorem gives a formula for $\Psi \left(i/k\right) +\gamma$ with transcendental terms. However, I suspect that proving that the result is transcendental is beyond the current state of the art. Perhaps it is implied by Schanuel's conjecture.

• Is this because we can use this digamma function in a type of construction of a basis for these guys? I was thinking that this might be the case after looking at the Lerch Transcendental but couldn't sort through all the details. Commented Jul 13, 2018 at 19:26
• In effort to pin down the relationship precisely: It looks like $\sum [a_0,a_1,\dots a_{k-1}] =-\frac{1}{k}\sum_{n=1}^k{a_{n-1}\Psi(\frac{n}{k})}$ Whenever the lefthandside converges. Commented Jul 15, 2018 at 3:34
• Which means we can use the mathematica code: a={1 ,0,0,-1,0}, Sum[(-1/Length[a])a[[n]]*PolyGamma[0,n/Length[a] ],{n,1,Length[a] }] to evaluate these guys. Commented Jul 15, 2018 at 3:51
• Also this seems pretty pointed related: sciencedirect.com/science/article/pii/S0022314X06002927 Commented Jul 15, 2018 at 3:53
• In fact... my statement might just be the same as their conjecture 6. Commented Jul 15, 2018 at 4:05

Extended comment:

Here is a way to "move the problem." After some elementary manipulation:

$$[a_0, a_1, \dots, a_{k-1}]=\sum_{n=1}^\infty{\frac{a_{n-1 \mod k}}{n}}=\int_{0}^1\frac{a_0x^0+a_1x+a_2x^2 +\dots+a_{k-1}x^{k-1}}{1-x^k}$$

Perhaps we can show that the righthand side is transcendental.

• Also... $$\gamma\sum_{n=1}^\infty{\frac{a_n}{n}}=\int_0^\infty \bigg( \sum_{n=1}^\infty a_ne^{-x^n} \bigg) \frac{dx}{x}$$ Commented Jan 24, 2020 at 18:43

In this paper it shown that this is the case that the sum is transcendental for all prime $k$.

Notes:

Somethings to be followed upon:

1) This post can be improved by finding out what happens for non prime $k$

2) The natural generalization of this:

For a whole number $s$ and vector with integer entries $[a_0, a_1, \dots, a_{k-1}]$ is it true that when $\sum_{n=1}^\infty{\frac{a_{n-1 \mod k}}{n^s}}$ is not divergent it limits to some transcendental number or zero? It looks to me like this is generally still undecided as Catalan's constant's algebraicity seems to undetermined (Infact we don't know if it's irrational or not).

$\sum_{n=1}^\infty\frac{a_n}{n} \in \mathbb{Q} \cup\mathbb{T}$.