# Convergence of series

• May 23rd 2012, 02:11 PM
Ant
Convergence of series
Hi,

$\sum_{n=1}^\infty \frac{n+2}{n^3 +1}$

I tried using the ratio test (ended up with 1 so wasn't helpful), similarly the nth root test and integral comparison test didn't work.

Can I say that:

$\frac{n+2}{n^3 +1} \leq \frac{1}{n^2}" alt="\frac{n+2}{n^3 +1} \leq \frac{1}{n^2}" />

and 1/n^2 converges by the integral comparison test, so by the comparison theorem, the sum converges?

Is the above correct? Thanks

edit: I couldn't get the Latex to work so I've taken off the tags, sorry!

• May 23rd 2012, 02:22 PM
skeeter
Re: Convergence of series
try the limit comparison test with the known convergent series $\sum \frac{1}{n^2}$
• May 23rd 2012, 02:27 PM
anonimnystefy
Re: Convergence of series
Hi

To use LaTeX put (TEX)(/TEX) tags around the text you want texified except that you shoul put [ instead of ( and ] instead of ) .
• May 23rd 2012, 02:35 PM
Ant
Re: Convergence of series
Thanks for the help. That's what I did, I just wasn't sure if it was right. Thank you!

When I put the tags around it comes up like this $\sum_{n=1}^\infty \frac{n+2}{n^3 +1}$

edit: it's working now! but I literally just copied and pasted form my original post! I'll try again.
• May 23rd 2012, 02:49 PM
Plato
Re: Convergence of series
Quote:

Originally Posted by Ant
Thanks for the help. That's what I did, I just wasn't sure if it was right. Thank you!

When I put the tags around it comes up like this $\sum_{n=1}^\infty \frac{n+2}{n^3 +1}$

edit: it's working now! but I literally just copied and pasted form my original post! I'll try again.

Simple comparison works.
$\sum_{n=1}^\infty \frac{n+2}{n^3 +1}\le \sum_{n=1}^\infty \frac{2}{n^2 }$