Loser66
  • Loser66
Find the radius of convergence: \(\sum_{n=0}^{\infty} z^{n!}\) Please, help
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
hold on
dan815
  • dan815
-1 to 1
Loser66
  • Loser66
how?

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dan815
  • dan815
lol im jk
dan815
  • dan815
hmmm greater than -1 and less than 1
dan815
  • dan815
how about this bound
dan815
  • dan815
|dw:1442626076018:dw|
Loser66
  • Loser66
\(\sum_{n=0}^\infty z^{n!}= \sum_{n=0}^\infty a_m z^{n!}\) where \(a_m = \begin{cases}0 ~~if ~~ m \neq n!\\m ~~if ~~m = n!\end{cases}\)
Loser66
  • Loser66
but I don't know how to deal with the first z, since \(sum_{n=0}^\infty z^{n!}= z^{0!} + z^{1!}+z^{2!}+...... = \color{red}{z} + \sum something we can handle\)
Loser66
  • Loser66
the red z kills me. hehehe ...
Loser66
  • Loser66
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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