Let $p,q$ be primes such that $3 \leq p < q$ and $\langle p, q \rangle = \lbrace mp+nq ~|~ m,n\in\Bbb{N} \rbrace$.
Compute the radius of convergence of : $$ S(z) = \sum_{s \in \langle p,q \rangle}{z^s}. $$
My first thought was to write $s$ as $mp+nq$ and then use the convergence test : $$ \frac{\lvert z^{(m+1)p+(n+1)q} \rvert}{\lvert z^{mp+nq} \rvert} \longrightarrow \lvert z^{pq} \rvert. $$ So the radius is $1$. But I'm pretty sure that it's not correct to use the test on two variables like above.
$\langle p,q \rangle$ is a subset of the positive integers. For $|z| < 1$ $$ \sum_{s \in \langle p,q \rangle}{|z|^s} \le \sum_{n=1}^\infty |z|^n = \frac{1}{1-|z|} < \infty $$ so that the radius of convergence is at least one. On the other hand, $\langle p,q \rangle$ is an infinite set, so that the series diverges for $z=1$.
Therefore the radius of convergence is exactly one.
Your proof is not correct because $mp+nq$ and $(m+1)p+(n+1)q$ are not necessarily two consecutive exponents of the series.