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

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?

• Your idea is good. Take $a_n=( (-1)^n+1)n$, say.
– lulu
Feb 22, 2020 at 23:30

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.
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 $$n$$, $$a_n=0$$.
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.