# Convergence of p-series

1. Jan 21, 2012

### autre

1. The problem statement, all variables and given/known data

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

3. 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?

Last edited: Jan 21, 2012
2. Jan 21, 2012

### vela

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

3. Jan 21, 2012

### autre

Sorry, typo -- p<1.

4. Jan 21, 2012

### vela

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

5. Jan 21, 2012

### autre

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.

6. Jan 21, 2012

### vela

Staff Emeritus
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?