## SolomonZelman one year ago infinite geometric series convergence. recap

1. anonymous

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

2. Nnesha

,-,

3. SolomonZelman

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