Obviously, the function $1/\cos(x)$ is unbounded. There are singularities at $x_k =k\pi + \pi/2$. Are there natural numbers that come arbitrarily close to $x_k$? Or is there any other way to prove that the assumption $1/\cos(n)< C$ leads to a contradiction?

You may want to have a look at the answer to [this question](https://math.stackexchange.com/questions/73262/provingthatmnsqrt2isdenseinr). – Xander Henderson Nov 28 '17 at 17:03

This follows from [Weyl's Equidistribution Theorem](https://en.wikipedia.org/wiki/Equidistribution_theorem) and the irrationality of $\pi$. – lulu Nov 28 '17 at 17:04

Indeed that should solve the problem. But Dirichlet's approximation theorem (thanks to Xander) should also work. – Swanhild Bernstein Nov 28 '17 at 17:09
1 Answers
Let us consider the continued fraction of $\frac{\pi}{2}$, which is infinite since $\pi\not\in\mathbb{Q}$:
$$ \frac{\pi}{2}=\left[1; 1, 1, 3, 31, 1, 145, 1, 4, 2, 8, 1, 6, 1, 2, 3, 1, 4, 1, 5, 1, 41,\ldots\right] $$
There are infinite convergents of such continued fraction with an odd denominator:
$$ \frac{2}{1},\frac{11}{7},\frac{344}{219},\frac{51819}{32989},\frac{52174}{33215},\frac{260515}{165849},\ldots$$
and their numerators give natural numbers $n$ such that $\cos(n)$ is closer and closer to zero.
Indeed, from
$$ \leftq_n\frac{\pi}{2}p_n\right\leq \frac{1}{q_n} $$
and the Lipschitzcontinuity of $\cos$ we get
$$ \left\cos(p_n)\right=\left\cos(p_n)\cos(\pi q_n/2)\right\leq\frac{1}{q_n}\approx\frac{\pi}{2p_n} $$
so $\left\{\frac{1}{\cos(n)}\right\}_{n\geq 1}$ is unbounded since $\{p_n\}_{n\geq 1}$ is unbounded.
 338,356
 40
 353
 787

Because $\pi/4 = \arctan(1) = 11/3+1/51/7+1/9 + $ I can create such a sequence to approximate $\pi/2 $, where the denominator is odd. – Swanhild Bernstein Nov 28 '17 at 18:48

@SwanhildBernstein: This is not granted to provide values of n such that cos(n) is arbitrarily close to 0. You have to consider accurate rational approximations to make the trick work. – Jack D'Aurizio Nov 28 '17 at 18:50

No, but it gives the quotients $p_n/q_n  \pi/2$ getting smaller and smaller, if $p_n/q_n \pi/2< 1$ then $p_n  q_n\pi/2 < 1/q_n .$ – Swanhild Bernstein Nov 28 '17 at 18:56

@SwanhildBernstein: not enough to ensure $\cos(p_n)\to 0$, however. – Jack D'Aurizio Nov 28 '17 at 18:56


By considering the convergents of a continued fraction, ie the best rational approximations. – Jack D'Aurizio Nov 28 '17 at 18:59

To do what? Twice the numerators of the partial sums of Mengoli series still do not work. – Jack D'Aurizio Nov 28 '17 at 19:10

But I can use mean value theorem $\cos(p_n) = \cos(p_n)  \cos(q_n\pi/2)\leq \sin(\xi_n)p_nq_n\pi/2 \leq p_n  q_n\pi/2\leq 1/q_n.$ That need not to be fasted way, but it should work? – Swanhild Bernstein Nov 28 '17 at 19:15

@SwanhildBernstein: the last inequality does not hold for generic approximations, only if $p_n/q_n$ is a convergent of $\pi/2$. That's the relevance of the continued fraction shown above. – Jack D'Aurizio Nov 28 '17 at 22:41