Cauchy-Hadamard Theorem

Theorem
Let $\displaystyle \sum_{n \mathop = 0}^\infty a_nz^n$ be a power series, $a_n \in \C$.

Then the radius of convergence $R$ is given by:
 * $\dfrac 1 R = \limsup \left|{a_n}\right|^{1/n}$

where $\limsup$ denotes the limit superior.

In this context, it is understood that $\dfrac 1 0 = \infty$ and $\dfrac 1 \infty = 0$.

Proof
Let $L = \limsup \left|{a_n}\right|^{1/n}$.

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

$\forall r \in \left[{0 \,.\,.\, \dfrac 1 L}\right]$, we have $L < \dfrac 1 r$, which means we can find $\epsilon > 0$ such that $L + \epsilon < \dfrac 1 R$, which means $r < \dfrac 1 {L + \epsilon}$

Let $\tilde{r} = r \left({L + \epsilon}\right)$, we have $0 \leq \tilde{r} < 1$.

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

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

Thus:
 * $\forall n>>1: \left|{a_n r^n}\right| < \left({L + \epsilon}\right)^n r^n = \tilde{r}^n$

This means that the series $\displaystyle \sum_{n \mathop = 0}^\infty \left|{a_n r^n}\right|$ is convergent too.

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

Thus, by the definition of radius of convergence, we have $r \leq R$, and this holds for all $r \in \left[{0 \,.\,.\, \dfrac 1 L}\right]$, hence $\dfrac 1 L \leq R$.

Let now $\epsilon \in (0 \,.\,.\, L)$ be fixed.

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

for all $k \in \N$.

Thus
 * $\displaystyle \displaystyle \left|{a_{n_k} \cdot \frac 1 {\left({L + \epsilon}\right)^{n_k}}}\right| > 1$

for all $k \in \N$.

So the series
 * $\displaystyle \sum_{n \mathop = 0}^\infty a_n \left({\frac 1 {L + \epsilon}}\right)^n$

is not convergent, which means $R \leq \dfrac 1 {L + \epsilon}$.

This holds $\forall \epsilon > 0$, so we can say that $R \leq \dfrac 1 L$

Now we have $\dfrac 1 L \leq R \leq \dfrac 1 L$, which means $R = \dfrac 1 L$