So i'm pretty proud of myself, I did this proof even though it's not due until after the $\underline{five}$ $\underline{day}$ $\underline{weekend}$ that I have coming up, and I just wanted to run it by y'all to see if there is anyway I could refine it. Also, I did the proof by contradiction, does anyone know how you would do this one directly? I tried for a long time and had some good ideas but they never came full circle.
The general idea of my proof is to assume that the sequence does not converge and then construct a subsequence that has no convergent subsequence.
Proof: Assume for the sake of contradiction that the hypothesis holds and yet $\{a_n\}$ does not converge;
$\therefore \exists \epsilon>0$ s.t. $\forall N \in \mathbb{N}$ $\exists n>N$ s.t. $|a_n-L|>\epsilon$.
So let $N \in \mathbb{N}$, so $\exists r_1>N$ s.t. $|a_{r_1}-L|>\epsilon$
so then $\exists r_2>r_1$ s.t. $|a_{r_2}-L|>\epsilon$..
Continuing this way $\exists r_{i+1}>r_i$ s.t. $|a_{r_{i+1}}-L|>\epsilon$
Therefore $\{a_{r_i}\}$ is a subsequence of $\{a_n\}$ for which has no convergent subsequence, contradicting our hypothesis. Therefore $\{a_n\}$ converges.
Thanks everyone this website is great!