# Determining convergence/divergence of a series.

• Mar 31st 2013, 07:07 PM
jamesrb
Determining convergence/divergence of a series.
$\sum_{n=0}^{\infty}\frac{e^4^n}{e^3^n}$

This is a practice problem for a test. The test doesn't cover the comparison test, ratio test, root test or limit comparison test. I should be able to do this using the nth-term test, p-series, integral test or other basic methods of evaluating sequences and series. With that said, I don't think the integral test can be used here as the function is not decreasing. If we take the limit of the function we get:

$\lim_{x \to \infty}\frac{e^4^n}{e^3^n}=\frac{\infty}{\infty}$

Which is an indeterminate form. However, if I use L'Hopital's rule I don't think I will get anywhere:

$\lim_{x \to \infty}\frac{e^4^n}{e^3^n}=\frac{4e^4^n}{3e^3^n}= \frac{16e^4^n}{9e^3^n} ...$

And so on. No am I starting to get desperate. I think maybe I need to rewrite this. Or I have overlooked a simpler answer. Or my answer thus far is wrong.

Can I rewrite this as:
$\left (\frac{e^4}{e^3} \right )^n$

That would mean that:
$\left | r \right |\geq 1$

And so the series diverges.
• Mar 31st 2013, 07:34 PM
Gusbob
Re: Determining convergence/divergence of a series.
Quote:

Originally Posted by jamesrb
$\sum_{n=0}^{\infty}\frac{e^4^n}{e^3^n}$
Can I rewrite this as:
$\left (\frac{e^4}{e^3} \right )^n$

That would mean that:
$\left | r \right |\geq 1$

And so the series diverges.

Yes, you can do that.
• Apr 1st 2013, 12:13 AM
SworD
Re: Determining convergence/divergence of a series.
You should immediately see that this series doesn't even go to 0 as n goes to infinity. It fails that test so it has to diverge.
• Apr 1st 2013, 12:49 AM
hollywood
Re: Determining convergence/divergence of a series.
Yes, SworD is correct - the individual terms $\frac{e^{4n}}{e^{3n}}=e^n$ do not go to zero as n goes to infinity, so the series must diverge. That was my first thought.