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I want to show that $\lim_{n\rightarrow\infty}(a_n) = 0$ where $a_n = \tan(n) (\frac{1}{e})^{n}$

My argument is that: $\tan(n) = \frac{\sin(n)}{\cos(n)} \in \mathbb{R}$ for $n \in \mathbb{N}$ since cos(x) only has roots of the form $x = -\frac{\pi}{2} + \pi k , k \in \mathbb{Z}$.

The convergence of the other term is obvious.

Is it correct?

Symeof
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    What in that argument would go wrong if in $a_n = f(n)\bigl(\frac{1}{e}\bigr)^n$ we chose $f(n) = e^n$ instead of $f(n) = \tan (n)$? – Daniel Fischer Aug 19 '15 at 14:22
  • I am not sure that I understand your point. The limit is clearly defined as f(n) = tan(n). I suppose you mean that this argument doesn't generalizes for f(n) an arbitrary function? I only need to prove this specific case.

    Or are you pointing at a direct flaw in my argument within the given definition of the problem?

     – Symeof Aug 19 '15 at 14:29
  • Actually the sequence $\tan n$ is highly irregular. I don't think that the limit $\lim_n \tan n / e^n$ does even exist. By the way, your argument is not correct, since you need to show that $\tan n$ is bounded, while your argument simply says that it is well defined using irrationality of $\pi$. – Crostul Aug 19 '15 at 14:29
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    Your argument only uses the fact that $\tan n \in \mathbb{R}$ for $n\in \mathbb{N}$. If the argument were correct, it would work for every function with that property. – Daniel Fischer Aug 19 '15 at 14:32
  • You are right, this doesn't suffice (a priori) to show that f(n) is bounded. The implicit idea behind my argument was that since $\cos n$ doesn't go towards 0, the ratio will be bounded by a real number. I now see that it is not necessarily true since there might be a way to always get closer to $\cos n = 0$ with bigger $n$. So the convergence isn't proven, although I suppose it shouldn't be too difficult to prove for $n \in \mathbb{N}$.

    Is my understanding correct?

     – Symeof Aug 19 '15 at 14:40
  • @Symeof Yes: now you understood where the matter is. In my opinion the proof that your sequence converges should be quite hard and tricky. Once you show that it converges, then it is quite obvious that the limit is $0$. – Crostul Aug 19 '15 at 14:44
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    @Crostul $\pi$ has an irrationality measure less than $8$, so $$\biggl\lvert \pi - \frac{2n}{2k+1}\biggr\rvert > \frac{c}{k^8}$$ for some $c$, hence $\bigl\lvert\bigl(k+\frac{1}{2}\bigr)\pi - n\bigr\rvert > \frac{c'}{k^7}$. If $\frac{2n}{2k+1}$ is close to $\pi$, then "$k\in \Theta(n)$", so we have $\lvert\cos n\rvert > \frac{c''}{n^{7}}$, whence $\lvert\tan n\rvert < K\cdot n^{7}$, so $e^{-n}\tan n \to 0$. – Daniel Fischer Aug 19 '15 at 15:06
  • @DanielFischer So would this fail if we replace $n$ with $\frac{L}{\pi} n$, where $L$ is a Liouville number? – Ian Aug 19 '15 at 15:22
  • @Ian Rather $\frac{\pi}{L}n$, or something similar. Then it may fail, not sure. That guarantees we have $n$ with $f(n) > n^k$ for every $k$, but the exponential may still win. – Daniel Fischer Aug 19 '15 at 15:31
  • @Symeof Why are the roots of the sine important to you? – principal-ideal-domain Aug 19 '15 at 15:56
  • @principal-ideal-domain : They don't matter indeed. Only those of cos(n). – Symeof Aug 19 '15 at 16:04

1 Answers1

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@Daniel Fischer already pointed out its relevance to the notion of irrationality measure.

This is related to the following question:

How fast $n$ gets closer to the zero set $\frac{\pi}{2} + \pi \Bbb{Z}$ of $\cos x = 0$?

We do not want a situation where (a subsequence of) $\cos n$ decays so fast that it can even beat the exponential factor $e^{-n}$. In other words, we want that $n$ stays moderately far from the zero set $\frac{\pi}{2} + \pi \Bbb{Z}$.

Related to this question is the irrationality measure of $1/\pi$. In particular, if the irrationality measure $\mu$ of $1/\pi$ is finite, then for each $\epsilon > 0$ there exists $c = c(\epsilon) > 0$ such that

$$ \forall q \in \Bbb{N}^{+}, p \in \Bbb{Z}, \quad \left| \frac{1}{\pi} - \frac{p}{q} \right| \geq \frac{c}{q^{\mu+\epsilon}}. $$

Now let us plug $q = 2n$ and $p = 2k+1$. Manipulating the inequality a little bit, we find that for some constant $c' = c'(\epsilon) > 0$,

$$ |n - (k+\tfrac{1}{2})\pi| \geq c' n^{-(\mu+\epsilon-1)}. $$

This implies that $\cos n$ stays away from $0$ in a predictable way: if $n$ is large, then

$$ |\cos n| \geq |\sin(c' n^{-(\mu+\epsilon-1)})| \geq \frac{2c'}{\pi} n^{-(\mu+\epsilon-1)}. $$

So it follows that

$$ |e^{-n}\tan n| \leq C n^{\mu+\epsilon-1} e^{-n} \xrightarrow[n\to\infty]{ } 0. $$

Finally, it is proven that $\mu$ is indeed finite. Therefore $e^{-n} \tan n$ converges to $0$.

Sangchul Lee
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