Comparing Convergence: Exploring the p-Series Divergence for p<1

[autre]

Homework Statement

$\sum\frac{1}{n^{p}}$ converges for $p>1$ and diverges for $p<1$, $p\geq0$.

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 $p<1$. $\sum n^{p-1}$ ...but this series converges? Where did I go wrong?

 

[vela]

Homework Statement

$\sum\frac{1}{n^{p}}$ converges for $p>1$ and diverges for $p<1$, $p\geq0$.

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 $<1$. $\sum n^{p-1}$ ...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?