Let me first put the questions succinctly:

For any whole number $m$, can we explicitly define rationals $\{\alpha_n \}$ such that $$\sum_{n=0}^\infty \alpha_n =\beta(\frac{1}{m},\frac{1}{m})?$$

Now for some exposition:

To answer this question I suspect that we'll need to answer the question: For whole numbers $r$ and $m$, How many integers satisfy $x^m+y^m=r$.

A "circle" defined as $C_{m,r} = \big \{ x+yi\in \mathbb{C}: x^m+y^m=r^m \big \}$ has an interior area of $4\frac{\beta(\frac{1}{m},\frac{1}{m})}{2m}$. For a justification you can check out some background of this question which can be found here: Proving $\int_0^r{(r^m-x^m)^{1/m}dx}=\frac{\Gamma\left(\frac{1}{m}+1\right)\Gamma\left(\frac{1}{m}+1\right)}{\Gamma\left(\frac{2}{m}+1\right)}r^2$

Now we can exploit this for the case $m=2$ and ask the question: How many Guassian integers are there such that $z\in C_{m,r}$? To do this we will need define a function $\chi$. I suspect that I will want to define the $\chi_m$'s differently dependent on $m$ so for now we will just define $\chi_2$. We let

$$\chi_2(x) = \begin{cases} -1 & \quad \text{if } x \text{ is congruent to 3 mod 4}\\ 1 & \quad \text{if } n \text{ is congruent to 1 mod 4}\\ 0 & \quad \text{if } n \text{ is congruent to 0 mod 2} \end{cases}$$.

So then in the case that $m=2$, we find that $$|\{z: z\in C_{m,r^{1/m} }\}|= \sum_{d|r} \chi_m(d)$$ I will need to justify this clearly but at this point I will simply reference the video I've linked below.

Then we can say the total number of Gaussian integers which are on some circle with radius $\sqrt{t}$ as we range $t$ from 1 to $r^2$ should be an approximation of the area of the circle. Moreover as we let $r$ get large this is approximation should get better and better. That is, $$ \lim_{r\to \infty}\frac{1}{r^2}\sum_{t=1}^{r^2} |\{z: z\in C_{m,t^{1/m} }\}| = \pi $$

I will state the conclusion in terms of the area of $C_{m,r}$. As $r$ gets large, the sum of all of the cardinalities of $C_{m,r^{1/m}}$ will approach the area of the interior of $C_{m,r}$ times $r^2$. Which for the case $m=2$ is $4\frac{\beta(\frac{1}{m},\frac{1}{m})}{2m}=4\frac{\beta(\frac{1}{2},\frac{1}{2})}{4}=\pi$.

This all requires a little more rigor but here's a picture:

enter image description here

For each of the circles with radius $\sqrt{r}$ we ask how many integer pairs are on the circumference of the circle and for large $r$ we will say that the total number of integer pairs on the circles should be about $\pi r^2$.

That is, $$\lim_{r\to\infty}\sum_{t=1}^{r^2} \sum_{d|r^{1/m}}{}\chi_m(d)= \pi r^2$$ and it turns out after some hand waving that we arrive at the conclusion that (and this is only for the $m=2$ case. Note that we don't even have a definition of $\chi_m $ in general):

$$ \lim_{r\to\infty}\sum_{t=1}^{r^2} \frac{1}{r^2}\sum_{d|r^{1/m}}{}\chi_m(d)=\sum_{d\in \mathbb{N}}^{\infty}\chi_m(d)/d=\pi/4$$ Somewhat ironically, we have taken a very long route to what every high school calculus student (who is a math nerd) should know. After we evaluate our $\chi_2$ function we find:

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}=\pi/4$$ But the question should haunt these calculus students because... where is the circle in $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+ \dots$? It certainly haunted me. Calculus is somehow such a powerful tool in that it can show you the truth of this claim using Taylor series and geometric sums without elucidating any circles. You might get to the other side of this proof disappointed. Where is the circle? I wonder if this question haunted Leibniz...

But this approach where we count using concentric circles in the complex plane can elucidate the circle.

Here is my point: What's so special about $m=2$?

Can we use the same approach for $m \neq 2$? Because if we can answer the question what's the size of $\{ z\in C_{m,r^{1/m}}\cap \mathbb{Z}[i]\}$ perhaps through thinking through a clever selection of $\chi_m$ for every whole number $m$ we may be able to get an explicit series (of rationals) equal to the quantity $\beta(\frac{1}{m},\frac{1}{m})$. The image which would correspond to the image above for the $m=3$ appears below.

enter image description here

For the $m=3$ case we should expect $$ \sum_{r=1}^{\infty} \frac{1}{r^2}|\{ z\in C_{m,r^{1/m}}\cap \mathbb{Z}[i]\}|= \frac{2}{3}\beta(\frac{1}{3},\frac{1}{3})$$ because that's the area in the unit "circle," $|x|^3+|y|^3=1$. This area is $$\frac{2}{3}\beta(\frac{1}{3},\frac{1}{3}) \approx 3.533277500570899914627378999296877405143737076405115060253$$ This checks out intuitively: It should be larger than $\pi$ but smaller than $4$.

The last couple of posts I have made were inspired by the video by 3Blue1Brown: https://www.youtube.com/watch?v=NaL_Cb42WyY.

  • $\begingroup$ [19] T. Schneider, Zur Theorie der Abelschen Funktionen und Integrale, J. Reine Angew. Math. 183 (1941) 110–128. May be a good lead. They demonstrate that $\beta(a,b)$ is transcendental for rational $a,b $ $\endgroup$ – Mason Aug 29 '18 at 17:55

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