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?
 A: 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.
A: 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.
A: 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
$\lvert a_n\rvert\not\to\infty$
A: Yes, an unbounded sequence can have a convergent subsequence. 
As Weierstrass theorem implies that a bounded sequence always has a convergent subsequence, but it does not stop us from assuming that there can be some cases where unbounded sequence can also lead to some convergent subsequence.
For example- 
if we consider a sequence $S=(100,1,100,2,100,3,100,4,100,5,....)$, it is unbounded but there is a subsequence (consider only odd terms) i.e. $ S_n= (100,100,100.....)$ which is convergent. 
So, there may exist some subsequence that converges in an unbounded sequence, though this is not always the case.
