Limit Comparison/Comparison Test on Non-rational functions

chrischoi614
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Homework Statement


Either the Comparison Test or Limit Comparison Test can be used to determine whether the following series converge or diverge.
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

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 :)
 
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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.
 
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 don't know how to simplify it =\
 
chrischoi614 said:
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 don't know how to simplify it =\

Then show us the algebra you did to get that. It's not right.
 
no... i didnt get it wrong :S... i just put the terms together...
 
chrischoi614 said:
no... i didnt get it wrong :S... i just put the terms together...

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.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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