% This is lecture no. 20 of MH1101/: Calculus II.

\chapter{The Ratio \& Root Tests}%
\label{cha:the_ratio_&_root_tests}

The convergence tests outlined so far cannot be easily applied to series containing factorial terms or $n$-th powers.
For this, we will need the following tests.

\section{The Ratio Test}%
\label{sec:the_ratio_test}

\begin{thrm}
	Ratio Test.
	Let $\left\{a_n\right\}$ be a sequence and assume that the following limit exists:
	\[
	\rho = \lim_{n \to \infty}\left\lvert \frac{a_{n+1}}{a_n}\right\rvert
	.\]
	Then the following implications hold true:
	\begin{align*}
		\rho < 1 & \implies \sum a_n\ \text{converges absolutely}\\
		\rho >1 \lor \rho = \infty & \implies \sum a_n\ \text{diverges}\\
		\rho = 1 & \implies\ \text{Ratio test is inconclusive}
	\end{align*}
\end{thrm}
\begin{proof}
	First, we consider the case where $\rho < 1$.
	We choose a number $r$ such that:
	\[
	\rho < r < 1
	.\]
	Since $\ds{\left\lvert \frac{a_{n+1}}{a_n}\right\rvert}$ converges to $\rho$, there exists a number $N$ such that:
	\[
	\left\lvert \frac{a_{n+1}}{a_n}\right\rvert < r,\ \forall n>N
	.\]
	That is:
	\[
	\lvert a_{n+1}\rvert < r\lvert a_n\rvert,\ \forall n>N
	.\]
	In particular, we have:
	\begin{align*}
		\lvert a_{N+1}\rvert & < r\lvert a_N\rvert\\
		\lvert a_{N+2}\rvert & < r\lvert a_{N+1}\rvert < r^2\lvert a_N\rvert\\
				     &\ \,\vdots\\
		\lvert a_{N+k}\rvert & < r^{k}\lvert a_N\rvert.
	\end{align*}
	Therefore,
	\begin{align*}
		\sum_{n=N}^{\infty} \lvert a_n\rvert & = \sum_{k=0}^{\infty} \lvert a_{N+k}\rvert\\
						     & \leqslant \sum_{k=0}^{\infty} r^{k}\lvert a_N\rvert\\
						     & = \lvert a_N\rvert \sum_{k=0}^{\infty} r^{k}.
	\end{align*}
	The geometric series on the right hand side converges because $0<r<1$, and so the left hand side converges by the comparison test.
	Thus the series $\sum a_n$ converges absolutely.

	\bigbreak
	\noindent
	Secondly, we consider the case where $\rho >1\lor\rho = \infty$.
	We choose a number $r$ such that $1<r<\rho$.
	Arguing as before with inequalities reversed, we conclude that there exists an integer $N$ such that:
	\[
		\lvert a_{N+k}\rvert \geqslant r^{k}\lvert a_N\rvert,\ \forall k\geqslant 0
	.\]
	Since $r^{k}$ tends to $\infty$, we see that the term $a_{N+k}$ does not tend to 0.
	Thus, by the $n$-th term test for divergence, we can conclude that the series $\sum a_n$ is divergent.
\end{proof}
\begin{exmp}
	Consider the series $\ds{\sum_{n=1}^{\infty} n^2}$ and $\ds{\sum_{n=1}^{\infty} n^{-2}}$.
	Find $\rho$ for each.
	\begin{sol}
		Let $a_n = n^2$.
		Then:
		\begin{align*}
			\rho & = \lim_{n \to \infty}\left\lvert \frac{a_{n+1}}{a_n}\right\rvert\\
			     & = \lim_{n \to \infty}\frac{(n+1)^2}{n^2}\\
			     & = \lim_{n \to \infty} \frac{n^2+2n+1}{n^2}\\
			     & = \lim_{n \to \infty}\left(1+\frac{2}{n}+\frac{1}{n^2}\right)\\
			     & = 1.
		\end{align*}
		On the other hand, let $b_n = n^{-2}$.
		Then:
		\begin{align*}
			\rho & = \lim_{n \to \infty}\left\lvert \frac{b_{n+1}}{b_n}\right\rvert\\
			     & = \lim_{n \to \infty}\frac{n^2}{(n+1)^2}\\
			     & = \lim_{n \to \infty}\frac{n^2}{n^2+2n+1}\\
			     & = \lim_{n \to \infty}\frac{1}{1+\frac{2}{n}+\frac{1}{n^2}}\\
			     & = 1.
		\end{align*}
		Thus, $\rho=1$ in both cases. However, one series converges and one diverges, therefore, the ratio test is inconclusive when $\rho = 1$.
	\end{sol}
\end{exmp}
\begin{exmp}
	Using the ratio test, show that $\ds{\sum_{n=1}^{\infty} \frac{1}{n!}}$ converges.
	\begin{sol}
		We compute the limit $\rho$.
		Let $a_n = \ds{\frac{1}{n!}}$.
		Then:
		\[
			\frac{a_{n+1}}{a_n} = \frac{1}{(n+1)!}\cdot \frac{n!}{1} = \frac{1}{n+1}
		.\]
		Thus we have:
		\[
		\rho = \lim_{n \to \infty}\left\lvert \frac{a_{n+1}}{a_n}\right\rvert = \lim_{n \to \infty}\frac{1}{n+1} = 0
		.\]
		Since $\rho < 1 $, the series converges.
	\end{sol}
\end{exmp}
\begin{exmp}
	Apply the ratio test to determine if $\ds{\sum_{n=1}^{\infty} \frac{n^2}{2^{n}}}$ converges.
	\begin{sol}
		Let $\ds{a_n = \frac{n^2}{2^{n}}}$.
		We have:
		\begin{align*}
			\frac{a_{n+1}}{a_n} & = \frac{(n+1)^2}{2^{n+1}}\cdot \frac{2^{n}}{n^2}\\
					    & = \frac{1}{2}\left(\frac{n^2+2n+1}{n^2}\right)\\
					    & = \frac{1}{2}\left(1+\frac{2}{n}+\frac{1}{n^2}\right)\\
			\therefore\rho = \lim_{n \to \infty}\left\lvert \frac{a_{n+1}}{a_n}\right\rvert & = \lim_{n \to \infty} \frac{1}{2}\left(1+\frac{2}{n}+\frac{1}{n^2}\right) = \frac{1}{2}.
		\end{align*}
		Since $\rho <1$, the series converges.
	\end{sol}
\end{exmp}

\section{The Root Test}%
\label{sec:the_root_test}

For some series, it is more convenient to use the following root test, based on the limit of the $n$-th roots $\root n\of{a_n}$ rather than the ratio of consecutive terms.

\begin{definition}
	Root Test.
	Let $\left\{a_n\right\}$ be a sequence and assume that the following limit exists:
	\[
		L = \lim_{n \to \infty}\root n\of{\lvert a_n\rvert}
	.\]
	Then the following implications hold true:
	\begin{align*}
		L < 1 & \implies \sum a_n\ \text{converges absolutely}\\
		L > 1 \lor L=\infty & \implies \sum a_n\ \text{diverges}\\
		L = 1 & \implies\ \text{Root test is inconclusive}
	\end{align*}
\end{definition}
\begin{proof}
	First, we consider the case where $L < 1$.
	We choose a number $r$ such that:
	\[
	L < r < 1
	.\]
	Now, because we have chosen $r$ such that $L<r$, there is some $N$ such that if $n\geqslant N$ we will have:
	\[
		\root n\of{\lvert a_n\rvert}< r\implies \lvert a_n\rvert<r^{n}
	.\]
	Now, the series:
	\[
	\sum_{n=1}^{\infty} r^{n}
	\]
	is a geometric series and because $0<r<1$ we in fact know that it is a convergent series.
	Thus, by the comparison test we conclude that:
	\[
	\sum_{n=N}^{\infty} \lvert a_n\rvert
	\]
	is convergent.
	However, since:
	\[
	\sum_{n=1}^{\infty} \lvert a_n\rvert = \sum_{n=1}^{N-1} \lvert a_n\rvert + \sum_{n=N}^{\infty} \lvert a_n\rvert
	,\]
	we know that the left hand side is convergent since it is the sum of two convergent sums.
	
	\bigbreak
	\noindent
	Next, we assume that $L>1$, and because of that, we know that there must be some $N$ such that if $n\geqslant N$ we have:
	\[
		\root n\of{\lvert a_n\rvert} > 1 \implies \lvert a_n\rvert > 1^{n}=1
	.\]
	However, if $\lvert a_n\rvert>1$ for all $n\geqslant N$ then we know that:
	\[
	\lim_{n \to \infty}\lvert a_n\rvert\neq 0 \implies \lim_{n \to \infty} a_n \neq 0
	.\]
	Therefore, the sum is divergent.
\end{proof}
\begin{exmp}
	Determine whether $\ds{\sum_{n=1}^{\infty} \left(\frac{n}{2n+3}\right)^{n}}$ converges.
	\begin{sol}
		Let $a_n = \ds{\left(\frac{n}{2n+3}\right)^{n}}$.
		Then:
		\[
			L = \lim_{n \to \infty}\root n\of{\lvert a_n\rvert} = \lim_{n \to \infty}\frac{n}{2n+3} = \frac{1}{2}
		.\]
		Since $L<1$, the series converges.
	\end{sol}
\end{exmp}

