Does the p-Series Diverge for p<1?

[autre]

Homework Statement

\sum\frac{1}{n^{p}} converges for p&gt;1 and diverges for p&lt;1, p\geq0.

The Attempt at a Solution

(1) Diverges: I want to prove it diverges for 1-p and using the comparison test show it also diverges for p. \sum\frac{1}{n^{1-p}}=\sum\frac{1}{n^{1}n^{-p}}=\sum n^{p}/n=\sum n^{p-1} for p&lt;1. \sum n^{p-1} ...but this series converges? Where did I go wrong?

 

[vela]

Homework Statement

\sum\frac{1}{n^{p}} converges for p&gt;1 and diverges for p&lt;1, p\geq0.

The Attempt at a Solution

(1) Diverges: I want to prove it diverges for 1-p and using the comparison test show it also diverges for p. \sum\frac{1}{n^{1-p}}=\sum\frac{1}{n^{1}n^{-p}}=\sum n^{p}/n=\sum n^{p-1} for &lt;1. \sum n^{p-1} ...but this series converges? Where did I go wrong?

It's not very clear what you're thinking here. Can you elaborate a bit? What is less than 1?

 

[autre]

It's not very clear what you're thinking here. Can you elaborate a bit? What is less than 1?

Sorry, typo -- p<1.

 

[vela]

What exactly are you trying to prove? You refer to "it" but what exactly is "it"?

 

[autre]

What exactly are you trying to prove? You refer to "it" but what exactly is "it"?

Oh, sorry -- I want to prove that \sum\frac{1}{n^{p}} diverges if \sum\frac{1}{n^{1-p}} diverges using the Comparison Test.

 

[vela]

If you're using the comparison test, you need to show that
$$\frac{1}{n^p} \ge \frac{1}{n^{1-p}} = n^{p-1}$$Can you do that?