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I have to find upper limit and lower limit of such a sequence: $$x_n=1+n \sin\left(\frac{n\pi}{2}\right)$$

I am stucked since $\sin\left(\frac{n\pi}{2}\right)$ will change and have values -1,0,1 as n changes.

Does it mean that our upper limit is $+\infty$ and lower limit is $-\infty$? Or should I divide the function into subsequences and say that they may converge to $-\infty$, 1,$+\infty$ for $\sin\left(\frac{n\pi}{2}\right)$ having values -1,0,1, respectively?

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I interpret the term upper limit as limit superior.

Guide:

  • Yes, consider subsequence. Choose subsequence such that $\sin \left( \frac{n \pi}2 \right)$ is a known constant.

  • Think of when does $\sin \left( \frac{n \pi}2 \right)=1$ and when does $\sin \left( \frac{n \pi}2 \right)=-1$

  • $n = 4k+1$ where $k \in \mathbb{Z}$ is one possible subsequence that you want to consider.

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  • $\begingroup$ There would be 4subsequences: $n=4k+1$ for which sequence will have value of $1+4k+1=4k+2$; $n=4k+2$ and $n=4k$ for which sequence have value 1; $n=4k+3$ for which sequence will have value $1-4k-3=-4k-2$ ; Does it mean I have 3limits? upper/lower limits are $+/- \infty$? $\endgroup$ – Sally Mar 31 '18 at 8:08
  • $\begingroup$ Well, you can consider either $3$ or $4$ subsequences depending on how you partition the number. We don't say we have $3$ limits. Each subsequence has its own limits. Yes, you get the upper/lower limits right. $\endgroup$ – Siong Thye Goh Mar 31 '18 at 8:11
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Almost done:

Your sequence.$x_n$ is not bounded above and below.

Not bounded above;

Assume $B >0$, real, is an upper bound.

Consider the subsequence $x_{n_k}$, where

$n_k=1+4k, k =0,1,2,...$

$x_{n_k}= 1+ (1+4k)\cdot 1, k=0,1,2,.$

Archimedes:

There is a $k_0$ with $k_0 \gt (B-2)/4$.

For $k\ge k_0 $:

$x_{n_k} \ge B$, hence not bounded above .

Similarly not bounded below.

There are subsequences that converge to $\infty, 0, + \infty$.

They are?

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  • $\begingroup$ Yes I got that values through subsequences! $\endgroup$ – Sally Mar 31 '18 at 8:29
  • $\begingroup$ Sally.I guess you are done.:)) $\endgroup$ – Peter Szilas Mar 31 '18 at 8:31

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