Radius of convergence for $\sum^{\infty}_{n=0}(-z)^n$

At first glance , I though that the series

$$\sum^{\infty}_{n=0}(-z)^n$$

diverges, since the minus sign alternates by the value of the exponent. For odd-valued exponents, the partial sum is negative, and vice versa. So since the series appears as divergent, I would think that the radius of convergence is zero. But this does not seem to be the case.

If I want to calculate the radius of convergence for this series, I use the formula: $$\frac{1}{R}=\mathop {\lim \sup }\limits_{n \to + \infty } (|c_n|)^{\frac{1}{n}}$$

for the sequence $(-z)^n$

$$R=\frac{1}{\mathop {\lim \sup }\limits_{n \to + \infty } |(-z)|^{-\frac{1}{n}}}=\mathop {\lim \sup }\limits_{n \to + \infty } (z)^{\frac{1}{n}}=1$$

So since this radius of convergence is correct, why does the sequence appear as a divergent series (by the alternating sign)?

Write your power series as $$ \sum\limits_{n = 0}^\infty {( - 1)^n z^n } . $$ By the Cauchy–Hadamard formula the radius of convergence $R$ is $$ R = \frac{1}{{\mathop {\lim \sup }\limits_{n \to + \infty } \left| {( - 1)^n } \right|^{1/n} }} = \frac{1}{{\mathop {\lim \sup }\limits_{n \to + \infty } 1^{1/n} }} = \frac{1}{{\mathop {\lim \sup }\limits_{n \to + \infty } 1}} = \frac{1}{1} = 1. $$ Thus, the power series converges if $|z|<1$ and diverges if $|z|>1$.