I have the series $$ P_i = 1 = \bigg[1 + \frac{q}{p} + \bigg(\frac{q}{p}\bigg)^2 + \bigg(\frac{q}{p}\bigg)^3 + \cdots + \bigg(\frac{q}{p}\bigg)^{i-1} \bigg]P_1 $$
We have the boundary condition $$ P_N=1 $$
Apparently, if we assume $p\neq q$, we obtain $$ P_N = \frac{1-\bigg(\frac{q}{p}\bigg)^N}{1-\bigg(\frac{q}{p}\bigg)}P_1 $$
Could someone explain how the power series was simplified to this?
A geometric sum. It is easy to prove (e.g. by induction) that if $a\ne 1$:
$\begin{align*} \sum_{0 \le k \le n} a^k &= \frac{a^{n + 1} - 1}{a - 1} \end{align*}$
If $a = 1$, obviously:
$\begin{align*} \sum_{0 \le k \le n} a^k &= n + 1 \end{align*}$