deletions | additions       diff --git a/ratio.tex b/ratio.tex  index e69de29..6278894 100644  --- a/ratio.tex  +++ b/ratio.tex 
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\documentclass[12pt]{article}  \usepackage[utf8]{inputenc}  \usepackage{amsmath}  \title{\LaTeX}  \date{}  \begin{document}  Let:  \begin{align}  S=\sum^{\infty}_{n=1} n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}  \end{align}  If I use the Ratio Test to determine whether S converges, I need to   determine:  \begin{align}  \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right|  \end{align}    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|\:(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)^{(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\:+\: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}\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)}\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| \\  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)^{2}\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| \\  L=\lim_{n\to\infty}\left|\:\frac{25(n\:+\:1)}{16n}\:\right| \\  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{25\left(1\:+\:\frac{1}{n}\right)}{16}\:\right| \\  L=\left|\:\frac{25\left(1\:+\:\frac{1}{\infty}\right)}{16}\:\right| \\  L=\left|\:\frac{25\left(1\:+\:0\right)}{16}\:\right| \\  L=\left|\:\frac{25\left(1\right)}{16}\:\right| \\  L=\left|\:\frac{25}{16}\:\right| \\  L=\left[\frac{25}{16}\right]  \end{align}  If $L < 1$ : $S$ Converges\\  \vspace{1em}   If $L > 1$ or $\infty$: $S$ Diverges \\  \vspace{1em}   If $L = 1$: S Inconclusive \\  \vspace{1em}  Using the above answer, we know that $S$ \\  \vspace{1em}  $L=\frac{25}{16} > 1$ : $S$ Diverges \\  \end{document}