# Thread: Check my work on alternating series problem?

1. ## Check my work on alternating series problem?

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?

2. A series $\displaystyle \sum a_n$ convergent if $\displaystyle \sum |a_n|$ is convergent. This is known as absolute convergence.

If the property of absolute convergence does not hold, but the series is STILL convergent, then the series is conditionally convergent.

3. Enclose the whole exponent in braces for the desired effect.

( - 1 ) ^ { ( n + 3 ) } gives $\displaystyle (-1)^{(n+3)}$

Decreasing and Decreasing to Zero is enough for an alternating series.

Get rid of the "alternating" (usually absolute values does this) and test it again. Success is "absolute" convergence.