Cauchy-Hadamard Theorem

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

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

where $\limsup$ denotes the limit superior.

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

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

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:
 * $\forall r \in \left[{0 \,.\,.\, \dfrac 1 L}\right]: L < \dfrac 1 r$

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

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

Let $\tilde r = r \left({L + \epsilon}\right)$.

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:
 * $\forall n \gg 1: \left\lvert{a_n}\right\rvert^{1/n} < L + \epsilon$

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

This means that the series $\displaystyle \sum_{n \mathop = 0}^\infty \left|{a_n r^n}\right|$ is 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 $r \in \left[{0 \,.\,.\, \dfrac 1 L}\right]$.

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

Let now $\epsilon \in \left({0 \,.\,.\, L}\right)$ be fixed.

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

for all $k \in \N$.

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

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.

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.