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Suppose that there is a sequence $a_n$ with the below subsequnces:

a) $a_n = \frac{n+2+1}{n^2 + n + 1}$ when $n$ is odd.

b) $a_n = \frac{sinn}{n}$ when $n$ is mulptiple of $4$

c) $a_n = \frac{1}{n} - 1$ when $n$ is even but not multiple of $4$

Which are the cluster points? The solution is $A=\{-1,0,1\}$ but I don't understand why.

My solution:

a) $a_n = \frac{n+2+1}{n^2+n+1} = \frac{1 + \frac{2}{n} + \frac{1}{n}}{n+1+\frac{1}{n}} \rightarrow 0$

b) $ \frac{-1}{n} \leq \frac{sinn}{n} \leq \frac{1}{n} ... \rightarrow 0$

c) $ a_n = \frac{1}{n} - 1 \rightarrow -1$

So, $A = \{-1,0 \}$

How did they find the cluster point $1$ ?

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  • $\begingroup$ The subsequence c is $a_n = 1/n - n$ or $a_n = 1/n - 1$ like in your solution? $\endgroup$ – The Student Mar 23 at 16:29
  • $\begingroup$ I edited it. Thank you $\endgroup$ – Dimitris Dimitriadis Mar 23 at 16:30
  • $\begingroup$ The subsequence c converges to -1. So, why -1 isn't a cluster point ? $\endgroup$ – Dimitris Dimitriadis Mar 23 at 16:41
  • $\begingroup$ Sorry, is 1 instead of -1 $\endgroup$ – The Student Mar 23 at 16:42
  • $\begingroup$ Ohh ok !! I have the same opinion about this. It is an exercise of a book that gives this solution. I mean $A = \{-1,0,1\} $ $\endgroup$ – Dimitris Dimitriadis Mar 23 at 16:43

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