Range of "p" for Convergence of Series (-1^n)/(n^p)

[ABoul]

Homework Statement

find the range of possible values of the real number p for which the series (from 1 to infinity) of [(-1)^n]/[n^p] is

a. absolutely convergent
b. conditionally convergent
c. divergent

Homework Equations

D'Alembert's ratio test
Cauchy's integral test

The Attempt at a Solution

i don't even know where to start!

 

[Dick]

Start with the integral test. Look it up and tell us what it says. Then try to apply it.

 

[ABoul]

Start with the integral test. Look it up and tell us what it says. Then try to apply it.

how the hell do you go about integrating something with [(-1)^n] in it? that's probably the main problem for me...

 

[Dick]

When applying the integral test, you take the absolute value of the series terms. It's a test of absolute convergence.

 

[ABoul]

ok. so the integral test gives me [n^(1 - p)]/[1 - p]. if p is greater than 1, the above series (with absolute values) converges to 0. however, can i this apply this rule directly to the main series?

 

[Dick]

You didn't read the statement of the integral test. To test the series you integrate 1/x^p from 1 to infinity. And you can apply the test because for p>1, x^p is decreasing and the integral exists (because the antiderivative at infinity goes to 0). That means the series is absolutely convergent. Once you know that, the signs on the terms don't matter. It still converges.