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I happened accross this excercise:

Let $a_n$ be a bounded, divergent sequence. Proove that there are at least two subsequences converging to different limits.

As a first strategy I thought of doing this:

Let $b_n \rightarrow L$ be a convergent subsequence (B-W theorem). Choose $\epsilon$ such that $\forall N, \exists n>N$ so that $|a_n - L|>\epsilon$. Using this post, build a subsequence as follows:

$n_1 = \min(n>1 / |a_n-L|>\epsilon)$

$n_k = \min(n>n_{k-1}/ |a_n-L|>\epsilon)$

The set $a_{n_k}, k>=1$ is a subsequence of $a_n$, therefore bounded, therefore contains a convergent subsequence, which cannot converge to $L$ (first question: is this ok?)

Moreover I wondered if $sup$ and $inf$ of the sequence can be used, and what are their properties. Is the following reasoning correct?

Let $S_k = {a_n, n>=k}$, since all $S_k$ are bounded sequences, $sup(S_k)$ exists. Furthermore, since $S_1\supset S_2 \supset S_3 ...$ then $\sup(S_1)\geq \sup(S_2)\geq \sup(S_3) ...$. Therefore, the sequence $\sup(S_1), \sup(S_2)...$ is monotone and bounded (therefore convergent).

Third question, is it true that $\displaystyle L = \lim_{n\rightarrow \infty}\inf(S_n) = \lim_{n\rightarrow \infty}\sup(S_n) \Leftrightarrow a_n\rightarrow L$?

Thanks

Bumblebee
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Tarifazo
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    3RD Q: Yes. If $A=(a_n)n$ is a bounded real sequence then $\lim{n\to\infty}\sup_{m\ge n}a_m$is the largest limit of a subsequence of $A$ and $\lim_{n\to\infty}\inf_{m\ge n}a_m$ is the smallest limit of a subsequence of $A.$ And $A$ is convergent iff all subsequences of $A$ converge to the same limit – DanielWainfleet Nov 14 '21 at 17:33

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