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The question below has been asked many times here, but I need quick approach just to determine the sequence is properly divergent or not.

Question: Is the sequence $(x_n)$ where $x_n= n\sin n$ properly divergent? That is either $\lim(n\sin n)=+\infty$ or $\lim(n\sin n)=-\infty$?

My attempt:

Let $x_n=n\sin n$; then clearly $(x_n)$ is unbounded above. Hence it must have a properly divergent subsequence say $(x_{n_{k}})$ such that, $lim(x_{n_{k}})\rightarrow +\infty$.

Also, $(x_n)$ is unbounded below and hence it must have properly divergent subsequence say $(x_{m_{k}})$ such that $lim(x_{m_{k}})\rightarrow -\infty$.

Hence, given that the sequence $(x_n)$ has two subsequences tending towards different infinities, $(x_n)$ is not properly divergent.

Is my attempt correct? (Especially the part of existence of properly divergent subsequences? I did not constructed those subsequence, but I directly assume there existence. Is it fine?)

Please help me...

amWhy
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Akash Patalwanshi
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    The word "clearly" does not constitute a proof. How do you know the sequence is unbounded above? – Robert Israel Jun 24 '20 at 16:00
  • One way, which you have glossed over, is to show that $\sin(n)$ is dense in $[-1,1]$. Then there is an increasing subsequence with each term greater than $1/2$ and a decreasing subsequence with each term less than $-1/2$, which will give you the result. – Integrand Jun 24 '20 at 16:01
  • @RobertIsrael sir yes i know. But if such question asked in exam such that, we just need to determine it is properly divergent or not. Then is my attempt works? – Akash Patalwanshi Jun 24 '20 at 16:01
  • @ClementYung sir, no that doesn't answer my question. Because, I just need to determine whether or not given sequence is properly divergent or not, without constructing those subsequence. I had mentioned this in question. – Akash Patalwanshi Jun 24 '20 at 16:06
  • @AkashPatalwanshi how can you assume their existence without proving them? That's clearly invalid. – Clement Yung Jun 24 '20 at 16:07
  • @ClementYung Because we know that, "if $(x_n)$ is unbounded sequence then, there exists a properly divergent subsequence. – Akash Patalwanshi Jun 24 '20 at 16:09
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    And how do you know that $(x_n)$ is unbounded? – Clement Yung Jun 24 '20 at 16:13
  • @ClementYung sir , Robert Israel sir, yes i think we need to construct unbounded subsequence first. To show show given sequence is not bounded. Or is there is any approach possible? – Akash Patalwanshi Jun 24 '20 at 16:18
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    I doubt we can prove that the sequence is unbounded without constructing the subsequence, so no your proof is invalid. – Clement Yung Jun 24 '20 at 16:43

1 Answers1

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Assume towards contradiction that it properly diverges say to $ \infty$.

Then for some $N_1 \in \Bbb{N}$ the sequence $(x_n)_{n=N_1}^\infty$ is positive and increasing.
Consider $N_2 \in [(2N_1+1)π,(2N_1+2)π]$ such that $N_2\in \Bbb{N}$.

You get that $N_2>N_1$ but In that interval $\sin(N_2)<0$.

can you continue from here?