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This question is related to my previous question: Limit Superior

Now the theorem for the limit superior states that if $y>s^*$, then there is an $N \in \mathbb{N}$ such that $\forall n \ge N \implies s_n<y$

Now, I have the following 2 questions:

1) Is it true that there will be an monotonically decreasing subsequence of $s_n$, for $n \ge N$ converging to $s^*$? (it does seem to me that it is true by construction, but I just wanted to make sure)

2) When proving that $s^*$ is indeed the largest accumulation point of the sequence $s_n$, is it sufficient to show that only finitely many terms of the sequence $s_n$ are bigger than $s+\epsilon$ , for any $\epsilon>0$ ?

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Well lets take an example $s_{n} = (n - 1)/n$ if $n$ is even and $s_{n} = 1/n$ if $n$ is odd. In this case clearly $s_{*} = 0$ and $s^{*} = 1$. There is no decreasing subsequence which tends to $s^{*}$. So the answer to 1) is NO.

For 2) you need to show that for any $\epsilon > 0$, $s^{*} + \epsilon$ is not a limit point. This can be easily done by considering the number $a = s^{*} + \epsilon / 2$. Clearly $a > s^{*}$ and hence after a certain value of $n$ all the values of $s_{n}$ are less than $a$. Hence these values differ from $s^{*} + \epsilon$ by at least $\epsilon / 2$. Clearly in this case $s^{*} + \epsilon$ is not a limit point. You can now see that your reasoning here is correct.

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