Proving limit

**mremwo** · February 16th 2011, 09:42 PM

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

Here's what I did:

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

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

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

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

**roninpro** · February 16th 2011, 09:52 PM

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

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

**mremwo** · February 16th 2011, 11:01 PM

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.

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