Cauchy-Hadamard Theorem/Complex Case/Proof 2

Proof
Let $L = \limsup \cmod {a_n}^{1/n}$.

We will consider only the case $0 < L < \infty$, as the cases $L = 0$ and $L = \infty$ follow quite simply from this one.

We have that:
 * $\forall r \in \closedint 0 {\dfrac 1 L}: L < \dfrac 1 r$

Thus there exists $\epsilon \in \R_{> 0}$ such that:
 * $L + \epsilon < \dfrac 1 R$

and so:
 * $r < \dfrac 1 {L + \epsilon}$

Let $\tilde r = r \paren {L + \epsilon}$.

Then:
 * $0 \le \tilde r < 1$

This means that the geometric series $\displaystyle \sum_{n \mathop = 0}^\infty$ is convergent.

From the properties of the limit superior, we know that:
 * $\forall n \gg 1: \cmod {a_n}^{1/n} < L + \epsilon$

Thus:
 * $\forall n \gg 1: \cmod {a_n r^n} < \paren {L + \epsilon}^n r^n = \tilde r^n$

This means that the series $\displaystyle \sum_{n \mathop = 0}^\infty \cmod {a_n r^n}$ is also convergent.

Thus $\displaystyle \sum_{n \mathop = 0}^\infty a_n r^n$ is absolutely convergent and therefore convergent.

Thus, by the definition of radius of convergence, we have:
 * $r \le R$

which holds for all $r \in \closedint 0 {\dfrac 1 L}$.

Hence:
 * $\dfrac 1 L \le R$

Let now $\epsilon \in \openint 0 L$ be fixed.

Because of the properties of the limit superior, we can extract from the sequence of the formula a subsequence such that:
 * $\cmod {a_{n_k} }^{1 / n_k} > L - \epsilon$

for all $k \in \N$.

Thus:
 * $\displaystyle \cmod {a_{n_k} \cdot \frac 1 {\paren {L + \epsilon}^{n_k} } } > 1$

for all $k \in \N$.

So the series:
 * $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {\frac 1 {L + \epsilon} }^n$

is not convergent.

Thus:
 * $R \le \dfrac 1 {L + \epsilon}$

This holds for all $\epsilon \in \R_{>0}$.

So:
 * $R \le \dfrac 1 L$

Hence:
 * $\dfrac 1 L \le R \le \dfrac 1 L$

which means:
 * $R = \dfrac 1 L$

as required.