Prove the convergence/divergence of $\sum \limits_{k=1}^{\infty} \frac{\tan(k)}{k}$

Question

Can be easily proved that the following series onverges/diverges?

$$\sum_{k=1}^{\infty} \frac{\tan(k)}{k}$$

I'd really appreciate your support on this problem. I'm looking for some easy proof here. Thanks.

Why do you think this is true? Can you name any source? The series is not alternating, but the terms do change sign often, if I'm not mistaken. This does not look obvious to me. — , Jun 17 '12 at 13:30
For two consecutive integers at least one of them has $|\tan k|\ge C$ for $C=\tan(\pi/2-\arctan 1)$. This shows at least that this series cannot be absolutely convergent. — Martin Sleziak, Jun 17 '12 at 13:31
@Thomas: i used W|A and gave more values, and based upon results i concluded it's divergent. Do you think that i'm wrong? I can change the title, then. — user 1591719, Jun 17 '12 at 13:35
Honestly, I have no idea by now. (But would not consider this approach as reliable ;-) — , Jun 17 '12 at 13:36
@Thomas: anyway, i think it's better so. You're right, since i have no proof for the divergence excepting some empirical results using W|A. — user 1591719, Jun 17 '12 at 13:37
Here's link Chris was trying to post: http://www.wolframalpha.com/input/?i=sum%5fk=1%5e10000%20tan%28k%29%2fk — Martin Sleziak, Jun 17 '12 at 13:48
@Cocopuffs: i got your point. For instance, if i chose $x=\arctan(n)+2n\pi$ then when let n to go to infinity we get $\frac{1}{2\pi}$, but when choosing $x=-\arctan(n)+2n\pi$ then we get $-\frac{1}{2\pi}$. Then it's obvious that there is no limit. — user 1591719, Jun 17 '12 at 14:01
@Chris aren't you confusing limit of 'function' with limit of a sequence? — qoqosz, Jun 17 '12 at 14:05
@qoqosz: Cocopuffs was referring to the limit of sequence, isn't it? — user 1591719, Jun 17 '12 at 14:06
@Chris I don't think it's quite so easy, you should probably argue with the Diophantine approximations to $\pi$ — Cocopuffs, Jun 17 '12 at 14:08
@Chris yes, I don't understand what you mean by $x$, but your comment looked to me like you were choosing two sequences $(x_k)$ to prove that limit of a function $\tan x / x$ doesn't exist. — qoqosz, Jun 17 '12 at 14:08
This article http://goo.gl/zEfj0 claims to prove that the limit of $tan(n)/n$ does not exist — leonbloy, Jun 17 '12 at 14:34
@leonbloy: you mean the sequence or the series? If you mean the sequence, then i've already proved that. — user 1591719, Jun 17 '12 at 14:35
@Chris: You've not proved it, it's much more complicated than that — leonbloy, Jun 17 '12 at 15:48
@leonbloy: then prove that i'm wrong. I'm referring now at the sequence limit only. — user 1591719, Jun 17 '12 at 15:58
@Chris: If you think you've proved that (i really don't get it), then post it as an answer. It would be strange that such a simple proof exists (look at the papers I've cited). — leonbloy, Jun 17 '12 at 16:25
@leonbloy: consider that you have to calculate $\lim_{x\to\infty} \frac{\tan(x)}{x}$ and then choose two different subsequences $y_{n} = \arctan(n)+2nπ$ and $z_{n} = -\arctan(n)+2nπ$. Replace x by these subsequences and take limit to infinity and get 2 different limits as i typed above. Do you agree? The point is that $\lim_{x\to\infty} \frac{\tan(x)}{x}$ yields different limits according to different subsequences. Therefore, the limit cannot exist. — user 1591719, Jun 17 '12 at 16:42
@Chris: That says nothing about $\lim \tan(n)/n$ for integer $n$ — leonbloy, Jun 17 '12 at 16:47
@leonbloy: It's better to use *text* for italics, instead of relying on LaTeX. See here for more Markdown formatting commands. — Zev Chonoles, Jun 17 '12 at 18:21

leonbloy · Accepted Answer · 2016-06-13T22:10:48.270

A proof that the sequence $\frac{\tan(n)}{n}$ does not have a limit for $n\to \infty$ is given in this article _{(Sequential tangents, Sam Coskey)}. This, of course, implies that the series does not converge.

The proof, based on this paper by Rosenholtz (*), uses the continued fraction of $\pi/2$, and, essentially, it shows that it's possible to find a subsequence such that $\tan(n_k)$ is "big enough", by taking numerators of the truncated continued fraction ("convergents").

(*) "Tangent Sequences, World Records, π, and the Meaning of Life: Some Applications of Number Theory to Calculus", Ira Rosenholtz - Mathematics Magazine Vol. 72, No. 5 (Dec., 1999), pp. 367-376

score 23 · Answer 2 · edited Jun 17 '12 at 14:24

Let $\mu$ be the irrationality measure of $\pi^{-1}$. Then for $s < \mu$ given, we have sequences $(p_n)$ and $(q_n)$ of integers such that $0 < q_n \uparrow \infty$ and

$$\left| \frac{1}{\pi} - \frac{2p_n + 1}{2q_n} \right| \leq \frac{1}{q_n^{s}}.$$

Rearranging, we have

$$ \left| \left( q_n - \frac{\pi}{2} \right) - p_n \pi \right| \leq \frac{\pi}{q_n^{s-1}}.$$

This shows that

$$ \left|\tan q_n\right| = \left| \tan \left( \frac{\pi}{2} + \left( q_n - \frac{\pi}{2} \right) - p_n \pi \right) \right| \gg \frac{1}{\left| \left( q_n - \frac{\pi}{2} \right) - p_n \pi \right|} \gg q_n^{s-1}, $$

hence

$$ \left| \frac{\tan q_n}{q_n} \right| \geq C q_n^{s-2}.$$

Therefore the series diverges if $\mu > 2$. But as far as I know, there is no known result for lower bounds of $\mu$, and indeed we cannot exclude the possibility that $\mu = 2$.

p.s. Similar consideration shows that, for $r > s > \mu$ we have

$$ \left| \frac{\tan k}{k^{r}} \right| \leq \frac{C}{k^{r+1-s}}.$$

Thus if $r > \mu$, then

$$ \sum_{k=1}^{\infty} \frac{\tan k}{k^r} $$

converges absolutely!

Alas, the irrationality measure of $\pi^{-1}$ is likely to be $2$. Nevertheless, that's a nice argument. — D. Thomine, Jun 17 '12 at 14:59

Pedro · Answer 3 · 2012-06-17T17:20:01.480

7

A simle example would be the term $\tan 121 / 121$. Noting that $\dfrac{\pi}{2}+2 \cdot 19 \cdot \pi \approx 120,95131$ shows why the term is much larger relative to the others. Here you can see a plot and spot the rogues. Those are the first 20k terms.

enter image description here

Here's a much more interesting take, of the first 50k terms. Note the way they align.

enter image description here

Basically, what we're worried about is how close an integer can get to $$\dfrac{\pi}{2}+2 k \pi$$

and that is a though question to answer.

edited Jun 17 '12 at 17:20

answered Jun 17 '12 at 14:32

Pedro

122,002

One might be able to get by with just a density argument. – R.. GitHub STOP HELPING ICE Jun 17 '12 at 15:05
@R.. I object to the use of "just" [insert result here]. Just yesterday another user said "Just because any polynomial of $n$ degree has $n$ complex roots". If you have the argument, use it, but it is not "just" an argument. The notion of density is something quite important in analysis, amongst others. I have no tools to prove it, so I tried to transmit what my idea is to the OP. – Pedro Jun 17 '12 at 15:10
I wrote that as a comment rather than an answer because it's an idea for something one might try in order to develop a proof. By "just a density argument", I meant avoiding quantitative bounds on how close integers can get and instead using the density of integers mapped into the unit circle. – R.. GitHub STOP HELPING ICE Jun 17 '12 at 15:14
@R.. That's interesting. How would you carry out that mapping? – Pedro Jun 17 '12 at 15:16
You can think of it as reduction modulo 2pi, complex exponentiation, etc. – R.. GitHub STOP HELPING ICE Jun 17 '12 at 16:38
@R.. I'm not really versed in what you're talking about (I get the $\mod 2 pi i$ (but I only know about integer congruences) and about $e^{i x}$ mapping to unitary vectors in the unit circle, but maybe you can be more detailed/explicit, please? – Pedro Jun 17 '12 at 16:40
1

Consider $x \approx y$ if and only if $x-y$ is an integer multiple of $2\pi$. This defines an equivalence relation on the reals, and the equivalence classes are an additive group (also a ring) and each has a representative element in $[0,2\pi)$. As an additive group, it's isomorphic to the multiplicative group of the unit circle in the complex plane under the mapping $x \to e^{ix}$ (which is independent of the representative chosen). – R.. GitHub STOP HELPING ICE Jun 17 '12 at 16:45
@R.. Great. Thanks. – Pedro Jun 17 '12 at 16:47
Sorry it's been too long since I used latex... can't remember the right code for ~ in math... – R.. GitHub STOP HELPING ICE Jun 17 '12 at 16:48

Prove the convergence/divergence of $\sum \limits_{k=1}^{\infty} \frac{\tan(k)}{k}$

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