# Comparison test and Limit Comparison test for series

• Nov 24th 2010, 05:43 PM
witcon
Comparison test and Limit Comparison test for series
Hi, I don't know how to show that the series Attachment 19842 converges or diverges using the comparison test or limit comparison test. All the series so far have been rational.

• Nov 24th 2010, 05:48 PM
mr fantastic
Quote:

Originally Posted by witcon
Hi, I don't know how to show that the series Attachment 19842 converges or diverges using the comparison test or limit comparison test. All the series so far have been rational.

Note that $\displaystyle \sqrt{n^4 + 1} - n^2 = \frac{1}{\sqrt{n^4 + 1} + n^2 }$.
• Nov 24th 2010, 06:08 PM
witcon
What should I compare it with?
• Nov 24th 2010, 06:29 PM
Prove It
$\displaystyle \frac{1}{n^2}$
• Nov 24th 2010, 07:09 PM
mr fantastic
Quote:

Originally Posted by witcon
What should I compare it with?

You are expected to note that $\displaystyle \frac{1}{\sqrt{n^4 + 1} + n^2} < \frac{1}{\sqrt{n^4} + n^2}$ ....
• Nov 25th 2010, 12:54 AM
Prove It
Which is also $\displaystyle < \frac{1}{n^2}$...