If an infinite sequence diverges to infinity, does it mean that all of its infinite subsequences diverge to infinity? Prove.
If $a_n$ sequnce diverges to infinity, then:
$$(\forall \epsilon>0)(\exists N)(\forall n>N)(a_n>\epsilon)$$
Assume that $a_{{n}_{k}}$ is a subsequence of $a_n$ and it does not diverge to infinity, then
$$(\exists \epsilon>0)(\forall N)(\exists n>N)(a_n \le\epsilon)$$
Now, let $m = n_k$. Now, let's pick an arbitrary $N.$ and an $\epsilon$ satisfying the second statement Then, there exists such $m$ that $a_m > \epsilon$ and $a_m \le \epsilon$. Which is a contradiction. Therefore
all infinite subsequences of a sequence that diverges to infinity diverge to infinity.
Is my proof correct?
Best Answer
Your proof is correct. You can prove this also directly. Let $a_{n_k}$ be a subsequence. Since for all $\epsilon>0$ there exists an $N$ such that for all $n>N$ we have $a_n>\epsilon$. Take $n_k>N$ and so $a_{n_k}>\epsilon$. So $a_{n_k}\to \infty$ as well.