Proof Proving that aj=aj−1+aj−2 cannot converge to a finite limit writing question. I was trying this problem and was hoping for some feedback as I am not sure if I proved it or not... I am trying to figure out where I went wrong as I feel I didn't run a good argument.
For $\epsilon>0,\exists N:|a_j-a_{j+1}|<\epsilon$ for$j>N$.
$a_j=a_{j-1}+a_{j-2}$
$|a_j|=|a_{N+1}+a_{j-1}+a_{j-2}-a_{N+1}|$    I added zero in a unique way.
$|a_j|\leq|a_{N+1}|+|a_{j-1}+a_{j-2}-a_{N+1}|$  Triangle inequality
$|a_j|\leq|a_{N+1}|+|a_j-a_{N+1}|$  Substitute in $a_j$
for $j>N$, $|a_j-a_{N+1}| < \epsilon$
$|a_j|\leq|a_{N+1}|+\epsilon$
$|a_j|-|a_{N+1}|\leq\epsilon$ Since epsilon can't be negative and we know $|a_j|-|a_{N+1}|<0$, is there a contradiction?

No, you do not arrive at a contradiction. For example, it is true that $|17|-|42|\le 0.001$ even though the left hand side is negative.
Instead, let $\phi$ be the larger solution of $\phi^2=\phi+1$. Show that $\phi>1$. Show by induction that $a_n>c\phi^n$ provided you pick $c>0$ such that $a_1>c\phi$ and $a_2>c\phi^2$.
