Testing a series for convergence

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EDIT: The latex doesn't seem to be working at all... not exactly sure why this is. I can't delete the post, so, uh... never mind, I guess?
EDIT v. 2.0: Yeah, so copying and pasting images from LatexIt is apparently beyond me... Thanks for the help, razored! Right, so, any help anyone can give will be well appreciated!

Homework Statement


Determine whether the following series converges or diverges. If possible, determine the sum of the series exactly. Justify your answer with the proper series test.
\sum_{n=2}^\infty{\sqrt{n^3+3}-\sqrt{n^3-3}}

Homework Equations


Comparison test, integral test, root test, ratio test, etc.

The Attempt at a Solution


Multiplying by the conjugate
\sqrt{n^3+3}+\sqrt{n^3-3}
produces the series
\sum_{n=2}^\infty{\frac{6}{\sqrt{n^3+3}+\sqrt{n^3-3}}}.

My first inclination was to try and find a series which is clearly larger, but I'm having trouble doing that. In previous problems like this that I've seen, there's been an n in the denominator, allowing me to eliminate the terms with square roots and produce a series which is always greater.

In this case, though, I can't, so I gave the ratio test a try, with painful results:
\lim_{n\rightarrow\infty}\frac{\sqrt{(n+1)^3+3}+\sqrt{(n+1)^3-3}}{\sqrt{n^3+3}+\sqrt{n^3-3}}.
I tried expanding the cubic terms, making it a bit ridiculous.
\lim_{n\rightarrow\infty}\frac{\sqrt{n^3+3n^2+3n+4}+\sqrt{n^3+3n^2+3n-2}}{\sqrt{n^3+3}+\sqrt{n^3-3}}
 
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Your Work :

http://texify.com/img/%5CLARGE%5C%21%5Ctext%7BWork%20from%20poster%3A%20%7D%20%5C%5C%5Csum_%7Bn%3D2%7D%5E%5Cinfty%7B%5Csqrt%7Bn%5E3%2B3%7D-%5Csqrt%7Bn%5E3-3%7D%7D%20%5C%5C%5Csqrt%7Bn%5E3%2B3%7D%2B%5Csqrt%7Bn%5E3-3%7D%20%5C%5C%5Csum_%7Bn%3D2%7D%5E%5Cinfty%7B%5Cfrac%7B6%7D%7B%5Csqrt%7Bn%5E3%2B3%7D%2B%5Csqrt%7Bn%5E3-3%7D%7D%7D%5C%5C%5Clim_%7Bn%5Crightarrow%5Cinfty%7D%5Cfrac%7B%5Csqrt%7B%28n%2B1%29%5E3%2B3%7D%2B%5Cs%20%20qrt%7B%28n%2B1%29%5E3-3%7D%7D%7B%5Csqrt%7Bn%5E3%2B3%7D%2B%5Csqrt%7Bn%5E3-3%7D%7D%5C%5C%20%5Clim_%7Bn%5Crightarrow%5Cinfty%7D%5Cfrac%7B%5Csqrt%7Bn%5E3%2B3n%5E2%2B3n%2B4%20%20%7D%2B%5Csqrt%7Bn%5E3%2B3n%5E2%2B3n-2%7D%7D%7B%5Csqrt%7Bn%5E3%2B3%7D%2B%5Csqrt%7Bn%5E3-3%7D%7D.gif
 
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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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