Let $x_n$ be a an unbounded sequence of non-zero real numbers. Then
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it must have a convergent subsequence.
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it can not have a convergent subsequence.
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$\frac{1}{x_n}$ must have a convergent subsequence.
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$\frac{1}{x_n}$ can not have a convergent subsequence.
Well, 1. is not true as say $x_n=n\ \forall n$, 2. I am not sure, but I guess there may be an unbounded sequence which may have an convergent subsequence, 3. is true due to Bolzano-Weierstrass as it is bounded sequence, 4. is false.
Am I right? Please correct me if I am wrong anywhere. Thank you.
Best Answer
Consider the sequence: $x_n=\begin{cases} n& n=2k\\\frac1n &n=2k+1\end{cases}$ this sequence is unbounded, but has a convergent subsequence, and note that $u_n=\frac1{x_n}$ is also unbounded but has a convergent subsequence. Neither $x_n$ nor $u_n$ are convergent themselves.
What can be said true about $(3)$ is that if $x_n$ is unbounded then $\frac1{x_n}$ has a convergent subsequence, simply take $x_{n_k}$ to be a strictly increasing subsequence which is unbounded, and show that $\frac1{x_{n_k}}$ is a bounded sequence.
(And as you said, $(1)$ and $(4)$ are indeed false.)