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Hypothesis: Suppose that $\underset{n \rightarrow \infty}{\lim} |a_n| / |a_{n+1}| = R$.

Goal: Show that $\sum a_n z^n$ has radius of convergence $R$.

Attempt:

  1. The radius of convergence of $\sum a_n z^n$ can be expressed as

    $$ \frac{1}{\limsup |a_n|^{1/n}} $$

  2. From this we can derive

    $$ \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_{n+1} \right|^{1/n}} = \left( \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_{n+1} \right|} \right)^{1/n} = \left( \frac{\underset{n \rightarrow \infty}{\limsup}|a_{n}|}{\underset{n \rightarrow \infty}{\limsup} \left|a_{n+1} \right|/|a_{n}|} \right)^{1/n} $$

    so that then

    $$ \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} = \left( R \frac{1}{\underset{n \rightarrow \infty}{\limsup} |a_{n}|} \right)^{1/n} $$

    But at this point I'm not sure what to make of this subresult.

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  • $\begingroup$ You can't move the $n$-th root across the $\limsup$. The $n$ is bound in $\limsup\limits_{n\to\infty}$. $\endgroup$ – Daniel Fischer Mar 4 '14 at 20:51
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Use d'Alembert rule:

Let $z\in\mathbb C$.

$$ \frac{|a_{n+1} z^{n+1}|}{|a_nz^n|} \to \frac{|z|}{R} $$ so when $|z|<R$ there is convergence, and when $|z|>R$ there is divergence.

Hence, $R$ is the radius.

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  • $\begingroup$ This is making use of the ratio test for real sequences, no? $\endgroup$ – user125104 Mar 5 '14 at 14:05
  • $\begingroup$ @user125104 Absolute convergence implies convergence. $\endgroup$ – Wilson of Gordon Sep 12 '14 at 12:04

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