The original series (-1 is raised to the n+3 power. Couldn't figure out how to raise more than just the bracket with LaTeX):

$\displaystyle \sum_{3}^{inf}(-1)^(n+3)/(7+\sqrt[5]{n+11})$

I took everything except the (-1)^n+3 part, and called it bn. In order for the series to be convergent, bn must:

1. be decreasing

2. have a limit of 0 when n->inf

so, $\displaystyle bn = 1/(7+\sqrt[5]{n+11})$

bn+1 is less than bn, so it is obviously decreasing.

And the limit is obviously 0, because the denominator will grow while the numerator stays 1.

Am I correct? Also, what does it mean to be "absolute" and "conditionally" convergent? Which one is this?