# Does the sequence have a convergent subsequence?

I'm trying to figure out if the sequence $e^{(-n)^n}$ where n is a natural number has a convergent subsequence? It's in a past exam paper. I know that obviously I can't apply the Bolzano-Weirstrass theorem because its not a convergence sequence but im not sure how to test for a convergent subsequence if the original sequence is not convergent. Thanks in advance

• What are, conceptually, some of the simplest subsequences of a sequence $(a_n)$? – Daniel Fischer Jan 7 '16 at 15:07
• "I know that obviously I can't apply the Bolzano-Weirstrass theorem because its not a convergence sequence" - conceptually, that statement doesn't even make sense. If the sequence were convergent, then by definition all subsequences would be convergent (with the same limit). I recommend that you review this theorem. – Math1000 Jan 7 '16 at 16:02
• it has two "convergent" subquences : $(a_{2n+1})$ to $0$ and $(a_{2n})$ "to $+\infty$", from that you can find all the convergent subsequences – reuns Feb 17 '16 at 22:03

What happens if you look at odd values of $n$?
The subsequence of odd indices $(a_{2n+1})$ should converge to zero, since you only have negative values in the exponent.