Which is the sup and inf of ${\frac{1}{sin(n)}: n \in N}$

Asked Nov 11 '20 at 20:36

Active Nov 11 '20 at 21:31

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Considering that we know the $\sup$ and $\inf$ of $\sin(n)$, can we easily find the bounds of $\Bigl\lbrace {\frac{1}{\sin(n)}: n \in \Bbb N}\Bigr\rbrace$ or we have to follow a different way to find them?

My solution We know that $sup(sin(n))$ = 1 and $inf(sin(n))$= -1 and also know that $sin(n)$ is dense in $[-1,1]$, according to this if $\sin(n)$ is close to $0$ then $\frac{1}{\sin(n)}$ is close to infinity. So, the $\frac{1}{\sin(n)}$ is not bounded.

edited Nov 11 '20 at 21:21

asked Nov 11 '20 at 20:36

Dimitris Dimitriadis

$\sin(n)$ is never equal to zero. – hamam_Abdallah Nov 11 '20 at 20:45
@hamam_Abdallah If we say that if sin(n) tends to 0 then the fraction tends to infinity ? – Dimitris Dimitriadis Nov 11 '20 at 20:47
$\sin(n)$ has no limit when $n\to +\infty$. – hamam_Abdallah Nov 11 '20 at 20:49
@hamam_Abdallah yes, but for a $n \in N$, $sin(n)$ is close to $0$ – Dimitris Dimitriadis Nov 11 '20 at 20:51
$\sin(n) $ is close to zero when $ n$ is close to $ k\pi$. – hamam_Abdallah Nov 11 '20 at 20:52
perfect, so $1/sin(n)$ is close to infinity ? – Dimitris Dimitriadis Nov 11 '20 at 20:53
${\sin(n)}$ is dense in $[-1,1]$. – hamam_Abdallah Nov 11 '20 at 20:57
You have to prove that ${\sin n\mid n\in\bf N}$ is dense in $[-1,1]$. – Bernard Nov 11 '20 at 20:57
This thread has many great answers regarding density of $\sin(n)$: https://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1 – Nov 11 '20 at 21:06
@hamam_Abdallah so $\frac{1}{sin(n)}$ is not bounded – Dimitris Dimitriadis Nov 11 '20 at 21:13
@Bernard If we prove this what can we say about $\frac{1}{sin(n))}$ ? – Dimitris Dimitriadis Nov 11 '20 at 21:15
This implies that we have values of $n$ such that $0$ can approached by $\sin(n)$ as close as we please, so $\frac1{\sin(n)},\dots$ – Bernard Nov 11 '20 at 21:21
@Bernard tends to infinity??? :) – Dimitris Dimitriadis Nov 11 '20 at 21:22
In absolute value (because there are values of $n$ such that $\sin(n)<0 $, and others for which it is $>0$). – Bernard Nov 11 '20 at 21:25
@Bernard yeap, so the set is not bounded – Dimitris Dimitriadis Nov 11 '20 at 21:27
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@DimitrisDimitriadis: Πράγματι – Bernard Nov 11 '20 at 21:32
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Consider convergents of $\pi:\ \frac 31,\frac{22}7,\frac{333}{106},\frac{355}{113},\frac{103993}{33102},\dots$ Then $3\approx \pi$, $22\approx 7\pi$, $333\approx 106\pi$, $355\approx 113\pi$, $103993\approx 33102\pi\dots$. Set $a_n$ as a numerator of $n$-th convergent. Then $\lim\limits_{n\to\infty}\frac 1{|\sin a_n|}=\infty$. – Pavel R. Nov 11 '20 at 21:35

Which is the sup and inf of ${\frac{1}{sin(n)}: n \in N}$

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