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Is the following claim correct?

$Claim:$ If a sequence $(a_n)$ of real numbers has a convergent subsequence, then it must be bounded.

I intuitively feel that the answer is no since one could (possibly) construct some oscillating sequence/function that, say, on odd values of $n$ would dance in a bounded section of $\mathbb{R}$ and, say, on even values would diverge to $+\infty$. (Additionally, the Bolzano-Weierstrass theorem statement would have been stronger if the claim was true.)

Can someone possibly provide a counter-example?

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    $\begingroup$ Your idea is good. Take $a_n=( (-1)^n+1)n$, say. $\endgroup$ – lulu Feb 22 '20 at 23:30
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Of course it is false. Consider the sequence $(0,1,0,2,0,3,0,4,\ldots)$. It has a sequence that converges to $0$ but the sequence itself is unbounded.

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A simple example might be something like $$a_n = \left( (-1)^n+1\right)n$$

We see that for even $n$ the term $a_n=2n$, while for odd $n$, $a_n=0$.

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You are correct. In fact, take any unbounded sequence $a_n$ and any convergent sequence $b_n$ and let

$$ c_n = \begin{cases} a_{1/2(i+1)} & i=1,3,5... \\ b_{i/2} & i=2,4,6... \end{cases} $$

Then $c_n$ will have a convergent subsequence $c_{2n}$ but will still be unbounded.

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