Prove that: if $$lim_{n\to\infty} \ \frac{a_n}{n} = a>0 \ \ \ \ \ \ \ (1)$$ then $$lim_{n\to\infty} \ a_n = \infty \ \ \ \ \ \ \ (2)$$
My proof: Using a definition of limit, (1) yields: $$\forall\epsilon>0 \ \ \exists N \ \ \forall n>N \ \ \left|\frac{a_n}{n} - a\right|<\epsilon$$ which is equivalent to: $$\forall\epsilon>0 \ \ \exists N \ \ \forall n>N \ \ n(a-\epsilon)<a_n<n(a+\epsilon) \ \ \ \ \ \ \ (3)$$ To prove (2) wee need to show that $$\forall M \ \ \exists N \ \ \forall n>N \ \ a_n>M \ \ \ \ \ \ \ (4)$$ So, let $M$ be any positive real number. Then let's take $N=\frac{M}{a-\epsilon}$. From (3) we know that $a_n>n(a-\epsilon)>N(a-\epsilon)=M.$ So indeed (4) holds.
Is this proof correct?
Edit:
Ok, I rethought this once again and I think proof above is indeed wrong. So let's take a second try:
Using a definition of limit, (1) yields: $$\forall\epsilon>0 \ \ \exists N \ \ \forall n>N \ \ \left|\frac{a_n}{n} - a\right|<\epsilon$$ which is equivalent to: $$\forall\epsilon>0 \ \ \exists N \ \ \forall n>N \ \ n(a-\epsilon)<a_n<n(a+\epsilon) \ \ \ \ \ \ \ (3)$$ To prove (2) we need to show that $$\forall M \ \ \exists \tilde{N} \ \ \forall n>\tilde{N} \ \ a_n>M \ \ \ \ \ \ \ (4)$$ So, let $M$ and $\epsilon$ be any positive real numbers.
From (3) we know that it exists $N$ s.t. $\forall n>N \ \ a_n>n(a-\epsilon)$.
Now let's take $\tilde{N}=max\left(N,\lceil\frac{M}{a-\epsilon}\rceil\right)$. We will show that $\forall n>\tilde{N} \ \ a_n>M$.
We have now two cases: $N\geq\lceil\frac{M}{a-\epsilon}\rceil$ or $N<\lceil\frac{M}{a-\epsilon}\rceil$.
Let's consider the first case. We have from (3)
$$\forall n>\tilde{N}=N \ \ a_n>n(a-\epsilon)>N(a-\epsilon)\geq \lceil\frac{M}{a-\epsilon}\rceil(a-\epsilon)\geq M$$
Now let's consider the second case. Because $\lceil\frac{M}{a-\epsilon}\rceil>N$, it holds from (3) that
$$\forall n>\lceil\frac{M}{a-\epsilon}\rceil \ \ a_n>n(a-\epsilon)>\lceil\frac{M}{a-\epsilon}\rceil(a-\epsilon)=M$$
Yes it looks more or less fine to me.
More briefly you could say something along these lines: definitely $a_n>\frac{an}{2}$ and since $b_n=n$ grows to infinity, so does $\frac{an}{2}$ and so does $a_n$.