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Loser66

  • one year ago

\(\sum_{n=1}^\infty n^az^n\) 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.

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  1. Loser66
    • one year ago
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    This is my attempt: if z =1, the sum is \(\sum_{n=1}^\infty n^a\) converges if and only if a = -1

  2. anonymous
    • one year ago
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    That can't be right, \(\sum\limits_{n\ge1}\frac{1}{n}\) diverges...

  3. anonymous
    • one year ago
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    I think you meant to say \(a<-1\)?

  4. Loser66
    • one year ago
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    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?

  5. Loser66
    • one year ago
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    Hence combine the two, only a =-1 satisfies both, right?

  6. Loser66
    • one year ago
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    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?

  7. Loser66
    • one year ago
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    or \(-1\leq a<0\)?? can it be?

  8. anonymous
    • one year ago
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    I agree with your conclusion for the second series, but for \(z=1\) the series diverges for \(a\ge1\). Right, it should be \(-1\le a<0\).

  9. Loser66
    • one year ago
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    But on this interval, discuss the convergence of the sum is not easy (part b). I don't know how to argue.:(

  10. Loser66
    • one year ago
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    @chris00

  11. anonymous
    • one year ago
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    Any complex number that lies on the circle \(|z|=1\) with \(z\neq1\) can be written as \(z=e^{it}\), with \(0<t<2\pi\). So you have \[\sum_{n=1}^\infty n^az^n=\sum_{n=1}^\infty n^ae^{int}=\sum_{n=1}^\infty n^a\cos nt+i\sum_{n=1}^\infty n^a\sin nt\]That should make things easier to work with...

  12. anonymous
    • one year ago
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    For the series to converge, both the real and imaginary parts must also converge. Notice that \[|n^a\cos nt|=n^a|\cos nt|\le n^a~~\implies~~\sum_{n=1}^\infty n^a\cos nt\le\sum_{n=1}^\infty n^a\]

  13. anonymous
    • one year ago
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    The same can be said for the imaginary part.

  14. Loser66
    • one year ago
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    Thank you. I got it. :)

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