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Loser66

  • one year ago

Find the radius of convergence: \(\sum_{n=0}^{\infty} z^{n!}\) Please, help

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  1. anonymous
    • one year ago
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    hold on

  2. dan815
    • one year ago
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    -1 to 1

  3. Loser66
    • one year ago
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    how?

  4. dan815
    • one year ago
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    lol im jk

  5. dan815
    • one year ago
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    hmmm greater than -1 and less than 1

  6. dan815
    • one year ago
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    how about this bound

  7. dan815
    • one year ago
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    |dw:1442626076018:dw|

  8. Loser66
    • one year ago
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    \(\sum_{n=0}^\infty z^{n!}= \sum_{n=0}^\infty a_m z^{n!}\) where \(a_m = \begin{cases}0 ~~if ~~ m \neq n!\\m ~~if ~~m = n!\end{cases}\)

  9. Loser66
    • one year ago
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    but I don't know how to deal with the first z, since \(sum_{n=0}^\infty z^{n!}= z^{0!} + z^{1!}+z^{2!}+...... = \color{red}{z} + \sum something we can handle\)

  10. Loser66
    • one year ago
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    the red z kills me. hehehe ...

  11. Loser66
    • one year ago
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    if n = 1 to infinitive, that is we don't have that red z, then apply Dirichlet's test, we can have the radius of convergence is \(R= \dfrac{1}{lim sup \sqrt[m] {a_m}}=1\)

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