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?

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.
