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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.

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    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.2012-06-17
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    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.2012-06-17
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    @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.2012-06-17
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    Honestly, I have no idea by now. (But would not consider this approach as reliable ;-)2012-06-17
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    @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.2012-06-17
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    Does the sequence tan(k)/k even tend to zero?2012-06-17
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    Here's link Chris was trying to post: http://www.wolframalpha.com/input/?i=sum%5fk=1%5e10000%20tan%28k%29%2fk2012-06-17
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    @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.2012-06-17
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    @Chris aren't you confusing limit of 'function' with limit of a sequence?2012-06-17
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    @qoqosz: Cocopuffs was referring to the limit of sequence, isn't it?2012-06-17
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    @Chris I don't think it's quite so easy, you should probably argue with the Diophantine approximations to $\pi$2012-06-17
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    @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.2012-06-17
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    @qoqosz: right. I thought it would help that point.2012-06-17
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    ref: http://mathforum.org/kb/message.jspa?messageID=5274962012-06-17
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    This article http://goo.gl/zEfj0 claims to prove that the limit of $tan(n)/n$ does not exist2012-06-17
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    @leonbloy: you mean the sequence or the series? If you mean the sequence, then i've already proved that.2012-06-17
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    @Chris The sequence (and hence the series)2012-06-17
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    @Chris: You've not proved it, it's much more complicated than that2012-06-17
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    @leonbloy: then prove that i'm wrong. I'm referring now at the sequence limit only.2012-06-17
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    @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).2012-06-17
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    @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.2012-06-17
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    @Chris: That says *nothing* about $\lim \tan(n)/n$ for integer $n$2012-06-17
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    @leonbloy: right. Apparently not to much.2012-06-17
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    @leonbloy: It's better to use `*text*` for italics, instead of relying on LaTeX. [See here for more Markdown formatting commands](http://meta.stackoverflow.com/editing-help).2012-06-17

3 Answers 3

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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

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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!

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    Alas, the irrationality measure of $\pi^{-1}$ is likely to be $2$. Nevertheless, that's a nice argument.2012-06-17
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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.

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    One might be able to get by with just a density argument.2012-06-17
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    @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.2012-06-17
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    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.2012-06-17
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    @R.. That's interesting. How would you carry out that mapping?2012-06-17
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    You can think of it as reduction modulo 2pi, complex exponentiation, etc.2012-06-17
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    @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?2012-06-17
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    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).2012-06-17
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    @R.. Great. Thanks.2012-06-17
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    Sorry it's been too long since I used latex... can't remember the right code for ~ in math...2012-06-17