extinction process problem in branching process

Question

I am trying understand solution to the branching process extinction problem as under:-

"Suppose in a branching process the offspring distribution is as follows

$p_k$ = $pq^k$, $q=1-p$, $0<p<1, k=0,1,2,3, \ldots$ Discuss the probability of extinction"

In the solution the $\phi(s) = \sum_{j=0}^{\infty}pq^js^j$.

I am not clear as to how this results in $\frac{p}{1-qs}$. If we take out $p$, isn't the summation a geometric series with first term $1$, common ratio $qs$, so as to give $\frac{p.1((qs)^n-1)}{qs-1}$? The fact that answer is $\frac{p}{1-qs}$. would imply that $(qs)^n=0$. Why is that so? Finally the solution says

$\phi(s)=s \implies \frac{p}{1-qs}=s$ and therefore $s_0 = p/q$

Wouldnt solving above equation result in a quadratic equation with two solutions for $s$? How is the solution $p/q$. Request help understand the solution

Answer 1

$$\lim_{N \rightarrow \infty} \sum_{j=0}^N p(qs)^j=\lim_{N \rightarrow \infty} \frac{p(1-(qs)^{N+1})}{1-qs}=\frac{p}{1-qs}$$

Note that $\lim_{N \rightarrow \infty} (qs)^{N+1} =0$ if $|qs|<1$.

For a general geometric series,

$$\sum_{k=0}^{\infty}r^k = \frac{1}{1-r}$$ if $|r|<1$.

As for the quadratic equation, there are indeed $2$ roots, the other solution is $1$.

However, there is a result that says that the smallest nonnegative solution to $$\phi(s)=s$$

is the extinction probability.