deletions | additions       diff --git a/ratio.tex b/ratio.tex  index dd60adf..241b468 100644  --- a/ratio.tex  +++ b/ratio.tex 
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  What is the value of this limit?  \begin{align}  L&=\lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right| L\:&=\:\lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right|  \\[1em] L&=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2(n\:+\:1)\:+\:1)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2(n\:+\:1)\:+\:1)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right|  \\[1em] L&=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:2\:+\:1)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:2\:+\:1)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\cdot\left(\frac{5}{4}\right)^{-(2n\:+\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\cdot\left(\frac{5}{4}\right)^{-(2n\:+\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\cdot\left(\frac{5}{4}\right)^{(-2n\:-\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\cdot\left(\frac{5}{4}\right)^{(-2n\:-\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3\:-\:2n\:-\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3\:-\:2n\:-\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{2}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{2}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{25}{16}\right)\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{25}{16}\right)\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:\frac{25(n\:+\:1)}{16n}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:\frac{25(n\:+\:1)}{16n}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:\frac{25n\left(\frac{n}{n}\:+\:\frac{1}{n}\right)}{16n}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:\frac{25n\left(\frac{n}{n}\:+\:\frac{1}{n}\right)}{16n}\:\right|  \\[1em] L=\lim_{n\to\infty}\left|\:\frac{25\left(1\:+\:\frac{1}{n}\right)}{16}\:\right| L\:&=\:\lim_{n\to\infty}\left|\:\frac{25\left(1\:+\:\frac{1}{n}\right)}{16}\:\right|  \\[1em] L=\left|\:\frac{25\left(1\:+\:\frac{1}{\infty}\right)}{16}\:\right| L\:&=\:\left|\:\frac{25\left(1\:+\:\frac{1}{\infty}\right)}{16}\:\right|  \\[1em] L=\left|\:\frac{25\left(1\:+\:0\right)}{16}\:\right| L\:&=\:\left|\:\frac{25\left(1\:+\:0\right)}{16}\:\right|  \\[1em] L=\left|\:\frac{25\left(1\right)}{16}\:\right| L\:&=\:\left|\:\frac{25\left(1\right)}{16}\:\right|  \\[1em] L=\left|\:\frac{25}{16}\:\right| L\:&=\:\left|\:\frac{25}{16}\:\right|  \\[1em] L=\left[\frac{25}{16}\right] L\:&=\:\left[\frac{25}{16}\right]  \end{align}  If $L < 1$ : $S$ Converges\\  \vspace{1em}