Can I prove the order limit theorem with a counter example? In this mathematical text they are giving a proof for the first statement using a proof by contradiction that I perfectly understand, but I just don't understand why they did not just give this counterexample:
$a_n = \frac{1}{n} : n\in\mathbb{N}$
This sequence has the property that all of its elements are greater than zero, and it converges to 0. Since 0 is not less than zero, we get a contradiction.
 A: Your counterexample is not a counterexample. The statement in question says the following:

Let $a_n$ be a convergent sequence and $\lim_{n\to\infty}a_n=a$. Then, if $a_n\geq 0$ for all $n$, then $a\geq 0$.

Your example merely shows that there exists some sequence $a_n$ such that all its values are positive, and the limit is $0$. The existence of one such sequence does nothing to prove the original statement. It is also not in conflict with the original statement, but it does not prove the original statement.
Also, note that the mere existence of any single sequence can never be proof enough of the original statement. This is because the original statement is talking about all sequences with a given property.
A: $a_n >0 ,\space \space \forall n>N $ and $a_n \to a $ $\implies a\ge 0$
Proof: Suppose, $a<0$
Then, $a< \frac{a}{2} <0$
Given, $a_n \to a$.
Then the interval $(\frac{3a}{2},\frac{a}{2})$ contains all but finitely many terms of $(a_n).$
This contradict the hypothesis that $a_n>0 \space \space \forall n>N$ .
In your case, $\frac{1}{n} >0$
And, $\lim (\frac{1}{n})=0\ge 0$
So, it is an example and doesn't contradict our statment.
