# Cluster points of a sequence $a_n$

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 $$1$$ ?

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