If $\lim a_n/n=a>0$ then $\lim a_n=\infty$ Prove that:
  if
  $$lim_{n\to\infty} \ \frac{a_n}{n} = a&gt;0 \ \ \ \ \ \ \ (1)$$
  then
  $$lim_{n\to\infty} \ a_n = \infty \ \ \ \ \ \ \ (2)$$


My proof:
Using a definition of limit, (1) yields:
$$\forall\epsilon&gt;0 \ \ \exists N \ \ \forall n&gt;N \ \ \left|\frac{a_n}{n} - a\right|&lt;\epsilon$$
which is equivalent to:
$$\forall\epsilon&gt;0 \ \ \exists N \ \ \forall n&gt;N \ \ n(a-\epsilon)&lt;a_n&lt;n(a+\epsilon) \ \ \ \ \ \ \ (3)$$
To prove (2) wee need to show that
$$\forall M \ \ \exists N \ \ \forall n&gt;N \ \ a_n&gt;M \ \ \ \ \ \ \ (4)$$
So, let $M$ be any positive real number. Then let's take $N=\frac{M}{a-\epsilon}$. From (3) we know that $a_n&gt;n(a-\epsilon)&gt;N(a-\epsilon)=M.$ 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:
$$\forall\epsilon&gt;0 \ \ \exists N \ \ \forall n&gt;N \ \ \left|\frac{a_n}{n} - a\right|&lt;\epsilon$$
which is equivalent to:
$$\forall\epsilon&gt;0 \ \ \exists N \ \ \forall n&gt;N \ \ n(a-\epsilon)&lt;a_n&lt;n(a+\epsilon) \ \ \ \ \ \ \ (3)$$
To prove (2) we need to show that
$$\forall M \ \ \exists \tilde{N} \ \ \forall n&gt;\tilde{N} \ \ a_n&gt;M \ \ \ \ \ \ \ (4)$$
So, let $M$ and $\epsilon$ be any positive real numbers.

From (3) we know that it exists $N$ s.t. $\forall n&gt;N \ \ a_n&gt;n(a-\epsilon)$.

Now let's take $\tilde{N}=max\left(N,\lceil\frac{M}{a-\epsilon}\rceil\right)$. We will show that $\forall n&gt;\tilde{N} \ \ a_n&gt;M$.

We have now two cases: $N\geq\lceil\frac{M}{a-\epsilon}\rceil$ or $N&lt;\lceil\frac{M}{a-\epsilon}\rceil$.

Let's consider the first case. We have from (3)

$$\forall n&gt;\tilde{N}=N \ \ a_n&gt;n(a-\epsilon)&gt;N(a-\epsilon)\geq \lceil\frac{M}{a-\epsilon}\rceil(a-\epsilon)\geq M$$

Now let's consider the second case. Because $\lceil\frac{M}{a-\epsilon}\rceil&gt;N$, it holds from (3) that

$$\forall n&gt;\lceil\frac{M}{a-\epsilon}\rceil \ \ a_n&gt;n(a-\epsilon)&gt;\lceil\frac{M}{a-\epsilon}\rceil(a-\epsilon)=M$$
No it is not correct. You did not define $ε$ in your proof. If you fix that you will also realize another error in your later reasoning.
Yes it looks more or less fine to me. 

More briefly you could say something along these lines: definitely $a_n&gt;\frac{an}{2}$ and since $b_n=n$ grows to infinity, so does $\frac{an}{2}$ and so does $a_n$.