# is my argument for $\lim_{n\rightarrow\infty}(a_n) = 0$ given $a_n = \tan(n) (\frac{1}{e})^{n}$ correct?

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?

• 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 $f(n) = \tan (n)$? Aug 19, 2015 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? Aug 19, 2015 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 $\pi$. Aug 19, 2015 at 14:29
• 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. Aug 19, 2015 at 14:32
• @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 $e^{-n}\tan n \to 0$. Aug 19, 2015 at 15:06

@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 $0$.

• Very nice! Thanks for your answer. Aug 19, 2015 at 15:36
• Great! We also have $\sum a_n$ converges. Aug 19, 2015 at 16:00