I'm trying to figure out how to do this proof, that if $x_n\to\infty$ then all its subsequences $x_{n_k}\to\infty$ as well.
I understand that there always exists a subsequence of a sequence that approaches infinity that approaches infinity, as you can just remove the first few values and arrive at the subsequence, but how do I prove all subsequences approach infinity?
Best Answer
The question in your title is mistaken: a divergent sequence may have convergent subsequences.
To the question in the main text. As always one needs to recall the definition of diverging to infinity: $x_n\to\infty$ if for all $M$, then $x_n\le M$ for only finitely many $n$. This entails that $x_{n_k}\le M$ for only finitely many $k$, so that $x_{n_k}\to\infty$