# Proving limit

• Feb 16th 2011, 09:42 PM
mremwo
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.
• Feb 16th 2011, 09:52 PM
roninpro
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 $\displaystyle a_n\to \infty$.
• Feb 16th 2011, 11:01 PM
mremwo
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.