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$$y_n=\rho^ny_0+(1+\rho+\rho^2+\cdots+\rho^{n-1})b.$$ 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?

Cookie
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user5826
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  • Do the know polynomial long division? – Eff Dec 21 '14 at 18:10
  • Just substitute numbers in your formula to assure that it is wrong. – Jihad Dec 21 '14 at 18:10
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    First of all, I suspect you are misreading your calculus book The value $\frac{1}{1-p}$ only equals the limit as $n\to\infty$. It can't equal the finite sum, since the finite sum obviously differs as $n$ differs... – Thomas Andrews Dec 21 '14 at 18:12

3 Answers3

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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 $p^n\to 0$.

Asaf Karagila
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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 $$ 1+p+p^2+\dots+p^{n-1}+\ldots=\frac{1}{1-p},\quad |p|<1. $$

Olivier Oloa
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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 $1-p$.

Tim Raczkowski
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