Question

infinite geometric series convergence.
recap

Answer 1

jeez solomon...you must be typing a book....a really long one..lol

Answer 2

,-,

Answer 3

\(\Large \color{blue}{ \displaystyle ^{\color{red}{~~~~~~~~~~~~~~~~~~~~~~~~~{\rm r}^4~~~~~+~~~~{\rm r}^3~~~~+~~~~{\rm r}^2~~~~+~~~~{\rm r}~~~~+~~~1}}_{\Huge _\text{_______________________________}}}\)
\(\large\color{blue}{ \displaystyle -{\rm r}+1{\huge|}~~-{\rm r}^5~~+~0{\rm r}^4~+~~0{\rm r}^3~~+~~0{\rm r}^2~+~~0{\rm r}~~+~1}\)
\(\large\color{red}{ \displaystyle -{\rm r}^5~~+~~{\rm r}^4 }\)
\(\large\color{blue}{ \displaystyle ^\text{____________} }\)
\(\large\color{red}{ \displaystyle -{\rm r}^4~~+~~0{\rm r}^3 }\)
\(\large\color{red}{ \displaystyle -{\rm r}^4~~+~~{\rm r}^3 }\)
\(\large\color{blue}{ \displaystyle ^\text{_____________} }\)
\(\large\color{red}{ \displaystyle -{\rm r}^3 ~~+~~0{\rm r}^2 }\)
\(\large\color{red}{ \displaystyle -{\rm r}^3 ~~+~~~{\rm r}^2 }\)
\(\large\color{blue}{ \displaystyle ^\text{_______________} }\)
\(\large\color{red}{ \displaystyle -{\rm r}^2 ~~+~~0{\rm r} }\)
\(\large\color{red}{ \displaystyle -{\rm r}^2 ~~+~~{\rm r} }\)
\(\large\color{blue}{ \displaystyle ^\text{______________} }\)
\(\large\color{red}{ \displaystyle -{\rm r}~+~1 }\)
\(\large\color{red}{ \displaystyle -{\rm r}~+~1 }\)
\(\large\color{blue}{ \displaystyle ^\text{___________} }\)
\(\large\color{red}{ \displaystyle 0 }\)


If you agree with (and understand) the above polynomial division, then you should get an intuitive understanding of why:

\(\color{black}{ \displaystyle \color{blue}{(-{\rm r}^{\rm n}+1)}\div \color{red}{(-{\rm r}+1)} ~~= \color{green}{{\rm r}^{{\rm n}-1}~+~{\rm r}^{{\rm n}-2}~+~....~+~{\rm r}^3~+~{\rm r}^2~+~{\rm r}~+~1} }\)

\((\)For all natural number n that are greater than 1 \()\)

Thus we get:

\(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\left(r^{k-1}\right)=1+r+r^2+r^3+...+r^{n-1} = \frac{-r^n+1}{-r+1}}\)

This is where: \(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\left(r^{k-1}\right)= \frac{1-r^n}{1-r}}\)

and \(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\color{orangered}{a_1}\left(r^{k-1}\right)= \frac{\color{orangered}{a_1}\left(1-r^n\right)}{1-r}}\) come from.
----------------------------------
Now, convergence of an infinite geometric series will be therefore determined by the convergence of the sequence of (1-r\(^n\))/(1-r)

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}(1+r+r^2+r^3+...+r^{n-1})=\lim_{n \rightarrow ~\infty}\left(\frac{1-r^n}{1-r}\right)}\)

after applying limit properties, we get:

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(\frac{1-r^n}{1-r}\right)=\left(\frac{1}{1-r}\right)\lim_{n \rightarrow ~\infty}\left(1-r^n\right) \\[1.9 em]
\large \displaystyle =\left(\frac{1}{1-r}\right)\left(1-\lim_{n \rightarrow ~\infty}r^n\right)}\)

\({\large \displaystyle =\left(\frac{1}{1-r}\right)-\left(\frac{1}{1-r}\right)\lim_{n \rightarrow ~\infty}r^n}\)

\(\scriptsize\color{ slate }{\scriptsize{\bbox[5pt, royalblue ,border:2px solid royalblue ]{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ }}}\)

So, \(r\ne1\) (because when r=1 we get an indetermine sum for the series)
And when r>1 the limit will go into infinity.
So 0>r>1 is so far verfied.

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}(r^n)}\)

for -1<r<0, the limit will approach zero (and thus exist) as well, and therefore by a convergence of this limit for ||r|<1, we verify the convergence of the sum of the series for |r|<1.