# Can a unbounded sequence have a convergent sub sequence?

I have been using the Bolzano-Weierstrass Theorem to show that a sequence has a convergent sub sequence by showing that it is bounded but does that mean that if a sequence is not bounded then it does not have a convergent sub sequence?

The sequence I am struggling is $((-1)^n \log(n))$ for all $n$ in the natural numbers. Now because the sequence is unbounded I am unsure how to prove whether it does or does not have a convergent sub sequence?

Take the sequence: $$(0,1,0,2,0,3,0,4,0,5,0,6,\cdots)$$It is unbounded and it has a convergent subsequence: $(0,0,0,\cdots)$. The Bolzano-Weierstrass theorem says that any bounded sequence has a subsequence which converges. This does not mean that an unbounded sequence can't have a convergent subsequence. What we can conclude is that any unbounded sequence has at least one unbounded subsequence.

• Is there any way to prove that it has a convergent subsequence though? even if at least one in unbounded or are there simply now sequences which wont have at least one convergent sub sequence? – Chris Oct 11 '15 at 22:02
• @Chris the sequence $(1,2,3,4, \dots)$ does not have a convergent sub sequence. – Hetebrij Oct 11 '15 at 22:03
• Yup, Herebrij beat me to it. I guess that's the simplest example. – Ivo Terek Oct 11 '15 at 22:04
• Ah yes of course, I am still not sure how I can prove whether or not my sequence has a convergent sub sequence though. – Chris Oct 11 '15 at 22:08

An unbounded sequence can have a convergent subsequence. An example is $a_n = n$ for n odd, $0$ for n even.

The correct contrapositive of Bolzano-Weierstrass is: a sequence with no convergent subsequence is unbounded.

As for $(-1)^n \log n$, consider any real x. For $n\geq e^{|x|+1}$, $|(-1)^n \log n - x|\geq 1$, so no subsequence of the sequence converges to x.

You can use the fact that if $\lvert a_n\rvert\to\infty$, then $(a_n)$ has no convergent subsequence:

If $(a_n)$ has a convergent subsequence, say $\displaystyle\lim_{k\to\infty}a_{n_k}=L$, then $\displaystyle\lim_{k\to\infty}\lvert a_{n_k}\rvert=|L|$ and therefore $\lvert a_n\rvert\not\to\infty$