# Value of $\limsup$ of $x_n=\sup\{x\in [0,x_{n-1}):\sin(1/x)=0\},n\ge 2$

A sequence is defined like the following $$x_1=1,$$ $$x_n=\sup\{x\in [0,x_{n-1}):\sin(1/x)=0\},n\ge 2$$ What is $\limsup x_n$?

I shall be highly obliged to those who will write the solution for me in detail.

Thank you, I am very weak in $\limsup$

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Well, I would write a quite explicit formula for $x_n$ if I were you: surely you know the zeroes of the sine function, hence... –  Did Jul 28 '12 at 21:46
I haven't written anything down, but if I calculated this right in my head it seems the sequence $x_n$ converges, so lim sup is the limit. –  Matt Jul 29 '12 at 0:02
If this is homework, you should tag it as such. –  Martin Argerami Jul 29 '12 at 4:21

Note that $\sin(\frac{1}{x}) = 0$ iff $x = \frac{1}{n \pi}$ for some $n \geq 1$.

Now consider the problem $\alpha_T = \sup \{x \in [0,T)\, | \ x \in \{ \frac{1}{n \pi} \}_n \} = \sup \{ \frac{1}{n \pi} \}_n \cap [0,T)$. Then we want to find the smallest $n\geq 1$ such that $\frac{1}{n \pi} < T \leq \frac{1}{(n-1) \pi}$ (ignoring the right hand upper bound if $n=1$). A small amount of work shows that $n = \lfloor \frac{1}{\pi T} \rfloor +1$. This gives $\alpha_T = \frac{1}{(\lfloor \frac{1}{\pi T} \rfloor +1)\pi}$. So, we have $x_2 = \alpha_1$, and $x_{n+1} = \alpha_{x_n}$.

It is easy to see that $x_2 = \alpha_1 = \frac{1}{\pi}$, and that if we assume $x_n = \frac{1}{n \pi}$, then $x_{n+1} = \alpha_{x_n} = \alpha_{\frac{1}{n \pi}} = \frac{1}{(n+1) \pi}$.

Consequently, we have $x_n = \frac{1}{n \pi}$. Hence $\limsup_n x_n = \lim_n x_n = 0$.

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So $$x_1=1\,\,,\,\,x_2=\sup\{x\in [0,1)\;:\;\sin 1/x=0\}$$ But $$\sin 1/x=0\Longleftrightarrow \frac{1}{x}=k\pi\,\,,\,k\in\Bbb Z\Longleftrightarrow x=\frac{1}{k \pi}\,\,,\,k\in\Bbb N$$ since we're working with positive $\,x'\,$s , and thus the supremum in this case is a maximum: $$\,x_2=\frac{1}{\pi}$$

Mimic the above to obtain $$x_3=\frac{1}{2\pi}$$ And now develope this into a pattern...

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