# Testing a series for convergence

1. Apr 22, 2009

### a.s

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!

1. The problem statement, all variables and given/known data
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.
$$\sum_{n=2}^\infty{\sqrt{n^3+3}-\sqrt{n^3-3}}$$

2. Relevant equations
Comparison test, integral test, root test, ratio test, etc.

3. 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.
$$\lim_{n\rightarrow\infty}\frac{\sqrt{n^3+3n^2+3n+4}+\sqrt{n^3+3n^2+3n-2}}{\sqrt{n^3+3}+\sqrt{n^3-3}}$$

Last edited: Apr 22, 2009
2. Apr 22, 2009