# Testing a series for convergence

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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