# Math Help - Comparison test and Limit Comparison test for series

1. ## Comparison test and Limit Comparison test for series

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

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

Note that $\displaystyle \sqrt{n^4 + 1} - n^2 = \frac{1}{\sqrt{n^4 + 1} + n^2 }$.
4. $\displaystyle \frac{1}{n^2}$
You are expected to note that $\displaystyle \frac{1}{\sqrt{n^4 + 1} + n^2} < \frac{1}{\sqrt{n^4} + n^2}$ ....
6. Which is also $\displaystyle < \frac{1}{n^2}$...