Does the Series with Terms 1/n^α Converge or Diverge?

[Dodobird]

Homework Statement

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.

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.

 

[micromass]

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?

 

[STEMucator]

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

 

[Dodobird]

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 got to prove something which is so new. I apologize for my bad English but I´m from abroad.

 

[Dodobird]

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

 

[STEMucator]

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 ).

 

[Dodobird]

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<br /> =2a_2+4a_4+...+N*a_N<br /> =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?

 

[Dodobird]

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

 

[Dodobird]

Okay well thanks for your help