Let's denote your answer by $\psi$.
To do it with series, let's compute the probability that you win on your $n^{th}$ turn. In order for that to happen, both you and your opponent must lose $n-1$ times. The probability of that happening is $(1-p)^{n-1}(1-q)^{n-1}$. After that, you must actually score, which happens with probability $p$. Thus the probability that you win on the $n^{th}$ trial is $$\psi_n=(1-p)^{n-1}(1-q)^{n-1}p$$ It follows that the probability that you win eventually is the sum of these, hence $$\sum_{n=1}^{\infty}\psi_n=\sum_{n=1}^{\infty}(1-p)^{n-1}(1-q)^{n-1}p=p \sum_{n=1}^{\infty}(1-p)^{n-1}(1-q)^{n-1}=p\sum_{n=0}^{\infty}(1-p)^{n}(1-q)^{n}$$ But this is just a geometric series, which we can sum to get $$\psi=\frac p{1-(1-p)(1-q)}=\frac p{p+q-pq}$$
Sanity checks: As the formula isn't exactly intuitive, it's worth thinking about some special cases. If, say, $p=1$ then this gives $\psi=1$ as it should. Also, if $q=0$, this correctly gives $\psi=1$ (ignoring the hopeless case in which both $p$ and $q$ are $0$). If $p=\frac 12 = q$ we get $\psi=\frac {\frac 12}{\frac 12+\frac 12-\frac 14}=\frac 23$ which may be recognized as the answer to the question "if two people alternate tosses of a fair coin, what is the probability that the first player gets the first $H$", so again the formula holds up.
