# Alternating Series Test

• Mar 26th 2011, 09:52 AM
VonNemo19
Alternating Series Test
Hi.

I've got a straight forward alternating series test problem:

Use the A.S.T. to determine whether the series

$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n\sqrt{n^3+1}}{n^5}$

converges.

We know that $\displaystyle \displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty }\frac{\sqrt{n^3+1}}{n^5}=0$, but when we try to show that

$\displaystyle a_{n+1}\leq{a_n}$

I am having trouble with this inequality. Help?
• Mar 26th 2011, 10:08 AM
Plato
Quote:

Originally Posted by VonNemo19
Use the A.S.T. to determine whether the series $\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n\sqrt{n^3+1}}{n^5}$ converges.
We know that $\displaystyle \displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty }\frac{\sqrt{n^3+1}}{n^5}=0$, but when we try to show that
$\displaystyle a_{n+1}\leq{a_n}$
I am having trouble with this inequality. Help?

Can you show that $\displaystyle f(x)=\dfrac{\sqrt{x^3+1}}{x^5}$ is a decreasing function?
• Mar 26th 2011, 10:11 AM
FernandoRevilla
An alternative, express:

$\displaystyle a_n=\sqrt{\dfrac{1}{n^7}+\dfrac{1}{n^{10}}}$

Inmediately we obtain $\displaystyle a_{n+1}\leq a_n$ .
• Mar 26th 2011, 03:37 PM
Connected
Quote:

Originally Posted by VonNemo19
We know that $\displaystyle \displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty }\frac{\sqrt{n^3+1}}{n^5}=0$

The limit includes $\displaystyle (-1)^n,$ and it tends to zero anyway since boundedness of $\displaystyle (-1)^n.$

You could also show that the series converges absolutely.