11
$\begingroup$

Question: $s\in\mathbb C$, Is $$\sum_{n=1}^\infty\frac{\cot \varphi\pi n}{n^s}$$ absolutely convergent, conditionally convergent or divergent, where $\varphi=\frac{1+\sqrt5}2$?

TL;DR, my progress
It is absolutely convergent if $\Re s>2$ and is divergent if $\Re s\le 1$. I don't know how to do when $\Re s\in(1,2]$.

Detailed progress
From Roth's theorem we can deduce that $\mu\left(\varphi\right)=2$, where $\mu$ denotes the irrationality measure.
If $\Re s\le 1$, the summand does not tend to $0$. (We can even find the limsup and the liminf of the summand when $s=1$.) Hence the sum diverges.
I have deleted my wrong "proof" of the convergence of the series when $\Re s>2$. I will fix it and add it to the question if I can. I believe there is only a little mistake in it and it can be fixed easily.


EDIT.
Numerical experimentPartial Sum of the SummandPartial Sum of the abs of Summand


EDIT 2.
Numerical calculation suggests $\sum_{n\le x}|\cot\varphi\pi n|=\Theta(x\log x)$. If we can prove this conjecture, we can obtain the absolute convergence of the series with condition $\Re s>1$ by using Abel's transformation.

$\endgroup$
  • $\begingroup$ With $\frac{1}{e^{-ix}-1}$ instead of $\cot(x)$ and $\Gamma(s) n^{-s} = \int_0^\infty t^{s-1}e^{-t}dt$ there is $\lim_{a \to 0}\sum_{m=1}^\infty \frac{1}{\textstyle e^t e^{(a+i \phi) m}-1}$ showing up but it is not obvious to tell how it converges in the sense of distributions (it should since for $\Re(s)$ larger than the irrationality measure its Mellin transform converges) and what is the behavior near $t=0$ needed for the analytic continuation. $\endgroup$ – reuns Apr 13 at 11:51
  • $\begingroup$ @ Kemono Chen Maybe it is not important but there is a flaw in your inequality chain. This inequality does not hold: $\left|\frac{\csc(\varphi\pi n-p\pi)}{n^s}\right| <\frac2\pi\frac1{|\varphi \pi n-p\pi|n^{\Re s}}$. Take $n=10$ and $p=16$ as an example. $\endgroup$ – Dr. Wolfgang Hintze Apr 23 at 12:22
  • $\begingroup$ @Dr.WolfgangHintze Sorry for that. I will try my best to find out my train of thought previously and fix it. $\endgroup$ – Kemono Chen Apr 23 at 12:55
4
$\begingroup$

My preliminary experiments suggest the conjecture that the sum is convergent for $s>1$.

Let us first have a look at some numerical results. Then we try to attack the problem anylytically.

I have added (15.04.) in §3 a possibly interesting stochastic approach which could lead to the proof of my conjecture.

17.04.19 $4 Structure of the partial sum

§1 empirical numerical results

Define the partial sum as

$$f(s,m) = \sum _{n=1}^m \frac{\cot (\pi n \phi )}{n^s}$$

and let

$$f(s) = \lim_{m\to\infty} f(s,m)$$

The following plots show the partial sum for a certain range of $m$ for the critical case $s=1$, for $=1.1$, and for $s=1.5$

enter image description here

For $s=1$ we see that the partial sums form strata of clouds, of finite width vertically as well as horizontally. The partial sums oscillate between the upper and the lower stratum. Hence the sum is divegent.

enter image description here

For $s=1.1$ the upper and lower bands of clusters tend to approach each other thus pointing to convergence.

enter image description here

Here, at $s=1.5$ convergence is obvious.

§2. Analytic approach

This is preliminary.

Writing

$$\frac{1}{n^s} = \frac{1}{\Gamma(s)}\int_0^\infty t^{s-1} e^{-n s}\,dt$$

and using the partial decomposition of the cotangent,

$$\cot(\pi x)= \frac{x}{\pi} \sum _{q=-\infty }^{\infty } \frac{1}{x^2-q^2}$$

we obtain for $f(s)$ a kernel

$$f_{\kappa}(s,n,t,q) = \frac{1}{\pi\Gamma(s) } t^{s-1} e^{-n t}\frac{x}{x^2-q^2}|_{x\to n \phi}$$

This kernel is to be summed over $n$ and $q$ and integrated over $t$. I don't know how far we can come on this path ...

Starting for example with the $n$-sum we have

$$f_{\kappa}(s,t,q) = \sum_{n=1}^\infty f_{\kappa}(s,n,t,q) \\ = \frac{e^{-t}}{2 \pi \phi } \left(\Phi \left(e^{-t},1,\frac{\phi -q}{\phi }\right)+\Phi \left(e^{-t},1,\frac{q+\phi }{\phi }\right)\right)$$

where

$$\Phi(z,s,a) = \sum_{k=0}^\infty \frac{ z^k}{(k+a)^s}$$

is the Hurwitz-Lerch transcendent.

§3. A stochastic analogy

Noticing that the numerator of the series exhibits a rather stochastic behaviour I found it worthwhile to study the following related problem:

Let $T$ be a continuous random variable with a given PDF $f(t)$ and define the (random) sum

$$g(s) = \sum_{n=1}^\infty \frac{t}{n^s}$$

We shall now use the three-series theorem of Kolmogorow /1/ to decide the convergence question which interestingly is given in the form "almost surely", meaning that the exceptions have probability zero /2/.

In order to see more clearly the relevance of this analogy we ask for the statistical properties of the numerator $u=\cot(v)$ with $v=\pi n \phi$, and $\phi$ the golden ratio.

It is clear that the value of $u$ depends only on the fractional part of $w=\{v\}$ of $v$. Now looking at the frequency distribution of $w$ and the first two moments are numerically close to $\frac{1}{2}$ (mean) and $\frac{1}{12}$ (variance), respectively, it seem reasonable to approximate $w$ by a true random variable $R$ with a PDF

$f_r(r) = \left\{ \begin{array}{ll} \frac{1}{\pi} & 0\le r\le \pi\\ 0& \text{else} \\ \end{array} \right. $

The distribution of $T = \cot(R)$ follows from $r = \text{arccot}(t)$ and $f(r) dr = f(r(t))(\frac{dr}{dt}) dt$ and results in the Cauchy distribution

$$f_{t}(t) = \frac{1}{\pi} \frac{1}{1+t^2}$$

The graph compares the experimental distribution of $u$ for $10^4$ consecutive values of $n$ with the Cauchy distribution.

enter image description here

Now the theorem of Kolmogoroff states

Let $(X_n)_{n\in \mathbb {N} }$ be independent random variables. The random series $\sum_{n=1}^{\infty}X_{n}$ converges almost surely in $\mathbb {R} $ if and only if the following conditions hold for some $A\gt0$:

(i) $\sum _{n=1}^{\infty }\mathbb {P} (|X_{n}|\geq A)$ converges.

(ii) Let $ Y_{n}=X_{n}\mathbf {1} \{|X_{n}|\leq A\}$ then $\sum _{n=1}^{\infty }\mathbb {E} [Y_{n}]$ , the series of expected values of $Y_{n}$ , converges.

(iii) $\sum _{n=1}^{\infty }\mathrm {var} (Y_{n})$ converges.

Application to our case:

Let $X_n = T n^{-s}$ then the variable $Y_n$ has the form $Y_n=T' n^{-s} $ where $T'$ has the trucated Distribution

$f_{t'}(t) = \left\{ \begin{array}{ll} \frac{1}{2 \text{arctan}(A)} \frac{1}{1+t^2} &|t|\lt A\\ 0& \text{else} \\ \end{array} \right. $

Hence for the probability in (i) we find easily

$\mathbb {P} (|X_{n}|\geq A) = \mathbb {P} (|T|\geq A n^s) = \frac{1}{n^s} \int_{A n^s}^{\infty } \frac{1}{\pi \left(t^2+1\right)} \, dt= n^{-s} \left(\frac{1}{2}-\frac{\text{arctan}\left(A n^s\right)}{\pi }\right)$

And since $\frac{1}{2}-\frac{\text{arctan}(z)}{\pi } \simeq \frac{1}{z \pi}$ for $z\to \infty$ we find

$\mathbb {P} (|X_{n}|\geq A) \simeq n^{-s} \frac{1}{A n^s \pi} = n^{-2s} \frac{1}{A \pi}$

Hence the sum (i) is convergent for $s\gt\frac{1}{2}$.

(ii) is trivial because $\mathbb {E} [Y_{n}]=0$ by symmetry.

(iii) the variance of the truncated distribution is

$\mathrm {var} (Y_{n})= \frac{1}{n^{2s}} 2 \int_0^{A n^s}\frac{1}{2 \text{arctan}(A)} \frac{t^2}{1+t^2}\,dt =\frac{n^{-2 s} \left(A n^s-\text{arctan}\left(A n^s\right)\right)}{\pi }$

For $n^s\to\infty$ this becomes $\mathrm {var} (Y_{n}) \simeq \frac{A}{\pi} \frac{1}{n^s}$ and convergence requires $s\gt 1$.

Summarizing: the random series $g$ is almost surely convergent if $s>1$. It is divergent if $s\leq 1$.

§4 Structure of the partial sums

This is one more of the several interesting questions related to the basic problem of the OP.

Here is the graph of a partial sum of $\cot(n \pi \phi)$

enter image description here

We observe two salient features

a) the change of the sign cuts the curve into pieces of incresing width

b) these pieces exhibit a self similar structure

First we try to find the position of the zeros of the curve. Close inspection of the interval depicted show that $p(1) < 0$, $p(2) > 0$, $p(3)\lt0$, $p(5) > 0$ and so on. We define a zero of the function to be the lower index of the sign change.

Hence the list of zeroes starts as follows $(1,2,3,5, 8, 13, 22, 35, 53, ...)$.

We recongnize this as the sequence of the well-known Fibonacci numbers, and we conjecture that this observation holds generally: the zeroes are located at the Fibanacci numbers $F(n)$.

In hindsight it is no wonder that Fibonacci numbers appear in a series employing the golden ratio which is approximated by the ratio $\phi \simeq F(n+1)/F(n)$.

To proceed we switch to the partial sum of the OP with $s=1$. Here's the graph again

enter image description here

We have already seen before that this partial sum remains bounded between two values. But it came as a surprise to me that the zeroes are in the same positions as in the previous partial sum.

We now plot the partial sum between two zeroes given by $n=F(k)-1$ and $n=F(k+1)$, and compare the cases $k=11$ and $k=13$ graphically

enter image description here

enter image description here

We observe that the curves have the same appearance, but in a different interval, and the structure of the curve has become more fine structured.

This process can be continued and gives us the complete picture of the partial sum up to the chosen maximum value $m$.

Specifically, the question of boundedness I raised already in a comment should find a confirming answer.

Notice for completeness that the same comparision can be made for the even Fibinacci rages which leads to the negative pieces of the partials sum.

For the time being I'll leave it to the peader to identfy the fractal structure (and determine e.g. the fractal dimension).

References

/1/ https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem

/2/ https://wikimedia.org/api/rest_v1/media/math/render/svg/232a94dbd97f2ce670d4987c5ad8ad82b072861b

$\endgroup$
  • 1
    $\begingroup$ Note $s$ is a complex number. The inequality $s>1$ is nonsense. $\endgroup$ – Jean-Claude Arbaut Apr 13 at 10:00
  • $\begingroup$ Not a particularly nice statement ... and wrong. $s>1$ makes complete sense in this context, and I'm sure you know that real numbers are a subset of the complex numbers. $\endgroup$ – Dr. Wolfgang Hintze Apr 13 at 10:52
  • $\begingroup$ I followed your step to do some other numerical experiment and I added it to the question body. Hope you don't mind. $\endgroup$ – Kemono Chen Apr 13 at 11:18
  • $\begingroup$ No, I don't mind, of course, but you could mention my Investigation. Your line drawing supplements my point drawing. $\endgroup$ – Dr. Wolfgang Hintze Apr 13 at 11:43
  • $\begingroup$ @ Kemono Chen: Thanks a lot. Your question brought to my attention a completly new field. I found it also interesting to study the picture if $\phi$ is replaced by other numbers linke $\pi$, $\sqrt{2}$, $e$ etc. While $\sqrt{2}$ leads to a similar fractal structure, $\pi$ results in a smooth behaviour. What if we replace $cot$ by $csc$ and so on? Fascinating! $\endgroup$ – Dr. Wolfgang Hintze Apr 13 at 12:22
1
$\begingroup$

This is an answer directly using the result proved by i707107 in this answer.
As shown by him or her, $$\sum_{k=1}^n|\cot\varphi\pi k|\sim\frac2\pi n\ln n.$$ If $\Re s>1$, by summing by parts (Abel's transformation) $$\left|\sum_{k=1}^n\frac{\cot\varphi\pi k}{k^s}\right|\le\sum_{k=1}^n\frac{|\cot\varphi\pi k|}{k^{\Re s}}\\ \sim \sum_{k=1}^n\frac{Ck\ln k}{k^{\Re s+1}}\to C\sum_{k=1}^\infty\frac{\ln k}{k^{\Re s}}$$ converges by the limit version of comparison test with $$\sum_{k=1}^\infty\frac{1}{k^{(\Re s-1)/2+1}}.$$ Then, combining with the proof of the convergence of the series while $\Re s\le 1$, we get the result:

The original series is absolutely convergent when $\Re s>1$ and is divergent when $\Re s\le 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.