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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. –  user20266 Jun 17 '12 at 13:30
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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. –  Martin Sleziak Jun 17 '12 at 13:31
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Does the sequence tan(k)/k even tend to zero? –  Cocopuffs Jun 17 '12 at 13:41
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This article goo.gl/zEfj0 claims to prove that the limit of $tan(n)/n$ does not exist –  leonbloy Jun 17 '12 at 14:34
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up vote 18 down vote accepted

A proof that the sequence $\frac{\tan(n)}{n}$ does not have a limit for $n\to \infty$ is given in this paper. This, of course, implies that the series is not convergent. The proof, is based on another paper by Rosenholtz, uses the continued fraction of $\pi/2$, and, essentially, 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").

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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. –  D. Thomine Jun 17 '12 at 14:59
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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. –  R.. 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 Tamaroff 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.. Jun 17 '12 at 15:14
    
@R.. That's interesting. How would you carry out that mapping? –  Pedro Tamaroff Jun 17 '12 at 15:16
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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). –  R.. Jun 17 '12 at 16:45
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