How to simplify this series?

Question

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?

@Gary, somewhat, although I've really only dealt with simplifying basic stuff like $\sum_{i=1}^N (\frac{1}{2})^i$. — user5965026, Apr 19 '20 at 13:21
Having $q/p$ in place of $1/2$ does not change much. Consider https://en.wikipedia.org/wiki/Geometric_progression#Derivation — Gary, Apr 19 '20 at 13:22

score 0 · Answer 1 · answered Apr 19 '20 at 14:38

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:

$\begin{align*} \sum_{0 \le k \le n} a^k &= n + 1 \end{align*}$

How to simplify this series?

1 Answers1