Comparison test and Limit Comparison test for series

**witcon** · November 24th 2010, 05:43 PM

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.

Thanks in advance.

**mr fantastic** · November 24th 2010, 05:48 PM

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.

Thanks in advance.

Note that $\displaystyle \sqrt{n^4 + 1} - n^2 = \frac{1}{\sqrt{n^4 + 1} + n^2 }$ .

**witcon** · November 24th 2010, 06:08 PM

What should I compare it with?

**Prove It** · November 24th 2010, 06:29 PM

$\displaystyle \frac{1}{n^2}$

**mr fantastic** · November 24th 2010, 07:09 PM

Originally Posted by witcon

What should I compare it with?

You are expected to note that $\displaystyle \frac{1}{\sqrt{n^4 + 1} + n^2} < \frac{1}{\sqrt{n^4} + n^2}$ ....

**Prove It** · November 25th 2010, 12:54 AM

Which is also $\displaystyle < \frac{1}{n^2}$ ...

Thread: Comparison test and Limit Comparison test for series