Ross: 8.9 "For all but finitely many..." I'm having trouble understanding this problem. 
Let $(s_n)$ be a sequence that converges.  A: Show that if $s_n \geq a$ for all but finitely many $n,$ then $\lim s_n \geq a$. 
In this case does "for all but finitely many" mean that only some terms of the sequence are greater than a? This is how I interpreted it and this is my proof:
Proof
Suppose $s_n \geq a$ for finitely many n, let $s=\lim s_n $. Then there exists an N such that $n>N$  implies 
$$|s_n - s| < a -s$$
Which implies 
$$|s_n|< a \Rightarrow s_n<a $$
Which holds since finitely many terms are bounded below by a, so infinitely many terms are bounded above by a.
Is my interpretation wrong? 
A: It means only finitely many terms are smaller than $a$, the rest of them are bigger than $a$.
So you should correct your proof as :
"Suppose $s_n \color{red}< a$ for finitely many $n$, then there exists an $N$ such that $n > N$ implies $s_n \geq a$."
Suppose $s< a$, try to find a contradiction.
A: Thank you all! I've been looking at proofs using contradiction but I want to be able to prove this another way. 
This is my next attempt, please correct me if I'm wrong:
Proof 
Suppose $(s_n)$ converges and $s_n \geq a$ for all but finitely many $n.$ Let $s=\lim s_n$.
There exists an N such that n>N implies 
$$|s_n -s | < s \Rightarrow |s_n|<2s $$
In particular for all n>N
$$s_n < 2s $$
Now since $a \leq s_n$ for all but finitely many $n,$ and since $s_n <2s$ for all $n>N,$ then $a \leq 2s$ which implies $a \leq s$.
