I am reading a proof that starts by assuming $\{t_{n}\}$ is a sequence satisfying $\sqrt{n}t_{n} \to k$, where $k$ is some finite constant. Later in the proof, the author says we can "relax" this assumption to $\sqrt{n}t_{n} = O(1)$.
I have two questions:
- Why are the two conditions $\sqrt{n}t_{n} \to k$ and $\sqrt{n}t_{n} = O(1)$ not equivalent?
- The author argues that if we assume $\sqrt{n}t_{n} = O(1)$, then for every subsequence $n'$ there exists a further subsequence $n''$ such that $\sqrt{n''}t_{n''} \to c$ for some finite constant $c$. Why is this true?
Some additional comments: I know that if $\sqrt{n}t_{n} = O(1)$, and if every subsubsequence $\{t_{n''}\}$ converges to the same limit $k$, then it must be that $\sqrt{n}t_{n} \to k$ and $\sqrt{n}t_{n} = O(1)$ are equivalent. Assumming the author is correct ---and that these are not equivalent--- it must be that $\sqrt{n}t_{n} = O(1)$ does not imply that every subsubsequence $\{t_{n''}\}$ converges to the same limit.
$\sqrt{n}t_{n} = O(1)$ means that the sequence $(\sqrt{n}t_{n})$ is bounded. That is weaker than convergent.
That is Bolzano -Weierstrass (a bounded sequence contains a convergent subsequence).