1. ## Proving limit

Show that if $\displaystyle lim(\frac{a_{n}}{n}) = L$ for $\displaystyle L>0$, then $\displaystyle lim(a_{n}) = + \infty$

Here's what I did:

$\displaystyle lim(\frac{a_{n}}{n}) = \frac{lim(a_{n})}{lim(n)} = L$

Then $\displaystyle \\ lim(a_{n}) = L \cdot lim(n)$

And since $\displaystyle \\ lim(n)$ goes to $\displaystyle +\infty$ and $\displaystyle L$ is a constant greater than 0,
the $\displaystyle lim(a_{n})$ also goes to $\displaystyle + \infty$

However, I'm pretty sure I need to use the definition of convergence to proves this, but I'm not sure how. Thank you.

2. What you have might serve as some intuition, but it does not work as a proof. You used the limit laws to rearrange the expressions, but they only work when the corresponding limits exist. Here is a slightly more formal argument:

Since $\displaystyle \lim_{n\to \infty} a_n/n = L$, we know that given any $\displaystyle \varepsilon>0$, when $\displaystyle n$ is sufficiently large, we have $\displaystyle |a_n/n-L|<\varepsilon$. Multiplying through by $\displaystyle n$ gives $\displaystyle |a_n-nL|<n\varepsilon$. Breaking the absolute values and rearranging, $\displaystyle n(L-\varepsilon)<a_n$.

This shows that as long as we pick $\displaystyle \varepsilon<L$ and take $\displaystyle n\to \infty$, we have $\displaystyle n(L-\varepsilon)\to \infty$, so $\displaystyle a_n\to \infty$.

3. Great, thanks! I had this all the way down to |a_n - nL| < ne but I always seem to forget that I am able to pick e to fit what I need so long as it is greater than 0.