Loser66
  • Loser66
\(\sum_{n=1}^\infty n^az^n\) for |z| =1, \(z\neq -1\) a) Which value of a , the series diverges at z =1 and converges at z = -1? b) discuss the convergence of the series with the values of a above when |z|=1 \(z\neq 1\) Please, help.
Mathematics
katieb
  • katieb
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Loser66
  • Loser66
This is my attempt: if z =1, the sum is \(\sum_{n=1}^\infty n^a\) converges if and only if a = -1
anonymous
  • anonymous
That can't be right, \(\sum\limits_{n\ge1}\frac{1}{n}\) diverges...
anonymous
  • anonymous
I think you meant to say \(a<-1\)?

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Loser66
  • Loser66
oh, yeah, I thought it diverges. if z = -1, the sum is \(\sum_{n=1}^\infty n^a (-1)^n\) converges for a <0, right?
Loser66
  • Loser66
Hence combine the two, only a =-1 satisfies both, right?
Loser66
  • Loser66
Condition for z = 1 and the sum is divergent is a =-1 condition for z =-1 and the sum is convergent is a <0 --------------------------------------------------- the condition of a which satisfies both is a =-1 , right?
Loser66
  • Loser66
or \(-1\leq a<0\)?? can it be?
anonymous
  • anonymous
I agree with your conclusion for the second series, but for \(z=1\) the series diverges for \(a\ge1\). Right, it should be \(-1\le a<0\).
Loser66
  • Loser66
But on this interval, discuss the convergence of the sum is not easy (part b). I don't know how to argue.:(
Loser66
  • Loser66
anonymous
  • anonymous
Any complex number that lies on the circle \(|z|=1\) with \(z\neq1\) can be written as \(z=e^{it}\), with \(0
anonymous
  • anonymous
For the series to converge, both the real and imaginary parts must also converge. Notice that \[|n^a\cos nt|=n^a|\cos nt|\le n^a~~\implies~~\sum_{n=1}^\infty n^a\cos nt\le\sum_{n=1}^\infty n^a\]
anonymous
  • anonymous
The same can be said for the imaginary part.
Loser66
  • Loser66
Thank you. I got it. :)

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