What Values of p Make the Series Converge Absolutely?

[GreenPrint]

Homework Statement

The sereis sigma[k=1,inf] [(-1)^k/k^p] converges conditionally for
(a) p<1
(b) 0<p<=1
(c) p>1
(d) p=0
(e) None of the above

Homework Equations

The Attempt at a Solution

The answer key said that (b) was the correct answer and I'm having trouble understanding why

sigma[k=1,inf] |(-1)^k/k^p| = sigma[k=1,inf] |1/k^p|

I got rid of the (-1)^k because the absolute value function will always make it positive and k^p will always be positive for k=1 to infinity so I just got ride of the absolute sign all together

sigma[k=1,inf] 1/k^p

I thought determine which values of p makes this series converge I could determine what values of the original series allows the series to converge absolutely

sigma[k=1,inf] 1/k^p

Is a p-series which converge whenever p is greater one by the integral test

I don't see were I'm going wrong thanks for any help

 

[muzak]

You missed the key words, "converges conditionally". When does it converge conditionally? What does that mean?

 

[GreenPrint]

ohhhh... that means when you take the absolute function around the function that describes the series what values of p does it diverge in which case it's 0<p<=1, I thought it asked me what values does it converge absolutley by accident thanks