Why does $1+p+p^2+\dotsb+p^{n-1}=\frac{1-p^n}{1-p}$

Question

$$y_n=\rho^ny_0+(1+\rho+\rho^2+\cdots+\rho^{n-1})b.$$ If $\rho \not=1$, we can write this solution in the more compact form $$y_n=\rho^ny_0+\frac{1-\rho^n}{1-\rho}b.$$

This is from Elem. Diff. Eq. - Boyce, DiPrima.

How was the more compact form derived? In my calculus text: $1+p+p^2+\dotsb+p^{n-1}=\frac1{1-p}$.
So where did Boyce/DiPrima get that numerator from?

Answer 1

The line you write is incorrect, $1+p+\ldots+p^{n-1}$ approaches to $\frac1{1-p}$ assuming that $|p|<1$.

Namely, if $|p|<1$, then $\lim\limits_{n\to\infty}(1+\ldots+p^n)=\dfrac1{1-p}$.

This is true exactly because $1+\ldots+p^{n-1}=\frac{1-p^n}{1-p}$, and when $|p|<1$, taking $n$ to infinity yields $p^n\to 0$.

Answer 2

An approach could be $$ (1-p)(1+p+p^2+\dots+p^{n-1})=(1+p+p^2+\dots+p^{n-1})-(p+p^2+\dots+p^{n-1}+p^n)=1-p^n $$ giving $$ 1+p+p^2+\dots+p^{n-1}=\frac{1-p^{n}}{1-p}, \quad p\neq1, $$ moreover if $|p|<1$, then $p^n \rightarrow 0$ and you get $$ 1+p+p^2+\dots+p^{n-1}+\ldots=\frac{1}{1-p},\quad |p|<1. $$

Answer 3

For $|p|<1$, the infinite sum $1+p+p^2+\dots=\frac{1}{1-p}$. The finite sum $1+p+\dots+p^{n-1}=\frac{1-p^n}{1-p}$. This can be easily seen by multiplying both sides by $1-p$.