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?
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You may want to have a look at the answer to this question. – Xander Henderson Nov 28 '17 at 17:03
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This follows from Weyl's Equidistribution Theorem and the irrationality of $\pi$. – lulu Nov 28 '17 at 17:04
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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
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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
$$ \left|q_n\frac{\pi}{2}-p_n\right|\leq \frac{1}{q_n} $$
and the Lipschitz-continuity 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.
Jack D'Aurizio
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Because $\pi/4 = \arctan(1) = 1-1/3+1/5-1/7+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
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@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
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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
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@SwanhildBernstein: not enough to ensure $\cos(p_n)\to 0$, however. – Jack D'Aurizio Nov 28 '17 at 18:56
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By considering the convergents of a continued fraction, ie the best rational approximations. – Jack D'Aurizio Nov 28 '17 at 18:59
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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
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But I can use mean value theorem $|\cos(p_n)| = |\cos(p_n) - \cos(q_n\pi/2)|\leq |\sin(\xi_n)||p_n-q_n\pi/2| \leq |p_n - q_n\pi/2|\leq 1/q_n.$ That need not to be fasted way, but it should work? – Swanhild Bernstein Nov 28 '17 at 19:15
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@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