Limit Comparison/Comparison Test on Non-rational functions

chrischoi614

1. The problem statement, all variables and given/known data
Either the Comparison Test or Limit Comparison Test can be used to determine whether the following series converge or diverge.
[i] which test you would use (CT or LCT)
[ii] which series you would use in the comparison.
[iii] does the series converge or not

The series of (root(n^4 +1) - n^2) n goes from 1 to infinity

2. Relevent equations
Series of 1/n^2? I am not too sure

3. The attempt at a solution

So what i did was drag out the n^2 from the root so it becomes (n^2)(root(1+(1/n^4)))
and I know i Think i have to compare this with 1/n^2 , I know this series converge, but however I do not know how to explain correctly, to compare it with 1/n^2, if 1/n^2 really is the right one to compare to, or should i be using limit comparison test? I am quite lost at the moment, I have tried everything, but the fact that all I can use is CT and LCT, I really don't know how to solve it. I know that root (n^4 + 1) is just really close to n^2, its that (+1) that make this series happen.... Pleasee and thanks :)

Dick

I would start by multiplying your expression by (sqrt(x^4+1)+n^2)/(sqrt(x^4+1)+n^2) and doing some algebra in the numerator. Then see what you think.

chrischoi614

I actually did that before, but i ended up with 2n^4 -(2n^2)(root(n^4 +1)) + 1 in the numerator, the fact that the square root is there really is making me struggle cus i dont know how to simplify it =\

Dick

Then show us the algebra you did to get that. It's not right.

chrischoi614

no..... i didnt get it wrong :S.... i just put the terms together...

Dick

I'm glad you are so confident but (sqrt(x^4+1)-n^2)*(sqrt(x^4+1)+n^2) doesn't have a sqrt in it if you expand it. Show us how you got 2n^4-(2n^2)(root(n^4 +1))+1 or we can't help you. Did you not change the sign on the n^2? That's the whole 'conjugate' thing that makes the sqrt cancel.