# Sequence Abel-Summation

1. Nov 10, 2012

### Dodobird

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

Show that the following sequence $\sum\limits_{n=1}^\infty \frac{1}{n^\alpha}$
for all real $\alpha > 1$ converges and for all real $\alpha \leq 1$
diverges.

3. The attempt at a solution
All I know is that the Abel-Summation is the only useful thing here, but I got no clue how to use it the right way and I heard that the common criteria won´t work.

I would be thankful for any hints or clues to get this proof running. Thank you in advance.

Last edited by a moderator: Feb 6, 2013
2. Nov 10, 2012

### micromass

Re: Sequence

First of all, this is a series and not a sequence.

Second. Have you seen things like the integral test?? Can you list all the tests which you've seen and think are useful?

3. Nov 10, 2012

### Zondrina

Re: Sequence

Break it up into cases. There are 3 cases for you to consider here.

4. Nov 10, 2012

### Dodobird

Re: Sequence

Thanks micromass and zondrina for your quick help

I´m sorry for the misconception. I meant series, sorry about that. Well our teacher told us/me that the quotient and/or root-test aren´t helpful here because the first one shows convergence and the second one divergence. He then talked about the summation by parts (Abel) and proved it with Weierstraß. I´m not sure if I´m allowed to use the integral-test on that because newer proofing techniques are forbidden until we have reached and proved them. I already got comfortable with sequences but the pace is so fast that we haven´t even got taught on series and now we gotta prove something which is so new. I apologize for my bad English but I´m from abroad.

5. Nov 10, 2012

### Dodobird

Re: Sequence

Is the integral test the same like the cauchy-criterion?

6. Nov 10, 2012

### Zondrina

Re: Sequence

The integral test allows you to test for the convergence of your series as long as f(n) is a monotone decreasing function ( Since you're starting at n=1 you don't have much to worry about ).

7. Nov 11, 2012

### Dodobird

Re: Sequence

Hi Zondrina and all the other helpers/readers. I used the Cauchy-Condensationtest:

For convergence:

$s_n = \sum\limits_{n=1}^\infty a_1+a_2+a_3+...+a_n$

with the estimate

$s_N=a_2+a_2+a_4+a_4+a_4+a_4+a_8+...+N*a_N =2a_2+4a_4+...+N*a_N =2(a_2+2a_4+...+ (\frac N2)*a_N)$

with

$N= 2^k$

$s_{2^k} =2(a_2)+2(a_4)+...+(2^{k-1})*a_k)=\sum\limits_{k=1}^n (a^k)a_{2^k}$

For divergence:

$t_n = \sum\limits_{k=1}^n \frac 1k$

With the comparison test
$s_n \leq t_n$

$\sum\limits_{k=1}^n (a^k)a_{2^k} \leq \sum\limits_{k=1}^n \frac 1k$

mit $a_{2^k} = \frac 1k$

The expression $\lim_{n \to \infty} \sum\limits_{k=1}^n 1$ goes to infinity, so $\sum\limits_{k=1}^n \frac 1k$ goes as well to infinity.

Is this correct???

8. Nov 12, 2012

### Dodobird

Re: Sequence

Would be cool if some1 could check it. If I´m close to the solution. Thank you.

9. Nov 13, 2012

### Dodobird

Re: Sequence

Okay well thanks for your help