We consider two powerful tests of convergence with many advantages.
The Ratio Test
Let $\displaystyle \sum_{n=1}^{\infty} a_n$ be a series with nonzero terms and let $$\rho=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|.$$ If $0 \leq \rho \lt 1,$ the series converges absolutely.
If $\rho \gt 1$ or $\rho=\infty,$ the series diverges.
If $\rho=1,$ the test is inconclusive.
Examples: Ratio Test
$\displaystyle \sum_{n=0}^{\infty} \frac{2^n+5}{3^n}$
$$
\begin{array}{lll}
\displaystyle \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|&=\displaystyle \lim_{n \rightarrow \infty} \left|\frac{\frac{2^{n+1}+5}{3^{n+1}}}{\frac{2^n+5}{3^n}}\right| &\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{2^{n+1}+5}{3^{n+1}}\cdot\frac{3^n}{2^n+5} &\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{2^{n+1}+5}{2^n+5}\cdot\frac{3^n}{3^{n+1}} &\mbox{}\\
&=\displaystyle \frac{2}{1}\cdot\frac{1}{3} &\mbox{}\\
&=\displaystyle \frac{2}{3} &\mbox{}\\
&\lt \displaystyle 1 &\mbox{}\\
\end{array}
$$
Thus, by the Ratio Test, the series converges absolutely.
$\displaystyle \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2}$
$$
\begin{array}{lll}
\displaystyle \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|&=\displaystyle \lim_{n \rightarrow \infty} \left|\frac{\frac{(2(n+1))!}{((n+1)!)^2}}{\frac{(2n)!}{(n!)^2}}\right| &\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{(2(n+1))!}{((n+1)!)^2}\cdot \frac{(n!)^2}{(2n)!} &\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{(2n+2)!}{(n!(n+1))^2} \cdot \frac{(n!)^2}{(2n)!} &\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{(2n)!(2n+1)(2n+2)}{(n!)^2(n+1)^2} \cdot \frac{(n!)^2}{(2n)!} &\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{(2n+1)(2n+2)}{(n+1)^2} \cdot \frac{1}{1} &\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{4n^2+6n+2}{n^2+2n+1} &\mbox{}\\
&=4&\\
&\gt 1\\
\end{array}
$$
Thus, by the Ratio Test, the series diverges.
$\displaystyle \sum_{n=1}^{\infty} \frac{\ln(n)}{n^2}$
$$
\begin{array}{lll}
\displaystyle \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|&=\displaystyle \lim_{n \rightarrow \infty} \left|\frac{\frac{\ln(n+1)}{(n+1)^2}}{\frac{\ln(n)}{n^2}}\right|&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{\ln(n+1)}{\ln(n)}\cdot \frac{n^2}{(n+1)^2}&\mbox{}\\
&=\displaystyle 1&\mbox{}\\
\end{array}
$$
Thus, the Ratio Test is inconclusive.
Special Note: Recall from last time that the Comparison Tests also didn't work well with this series.
We showed that the series converges using the Integral Test.
Special Note: Recall from last time that the Comparison Tests also didn't work well with this series.
We showed that the series converges using the Integral Test.
Pearl of Wisdom #1
The Ratio Test is often used for series involving factorials or exponentials.
Dire Warning #1
If $\displaystyle \sum_{n=1}^{\infty} a_n$ is a series of nonzero terms and $$\rho=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
The sum of the series $\displaystyle \sum_{n=1}^{\infty} a_n$ is NOT $\rho.$
The Root Test
Let $\displaystyle \sum_{n=1}^{\infty} a_n$ be a series with nonzero terms and let $$\rho=\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}.$$ If $0 \leq \rho \lt 1,$ the series converges absolutely.
If $\rho \gt 1$ or $\rho=\infty,$ the series diverges.
If $\rho=1,$ the test is inconclusive.
Examples: Root Test
$\displaystyle \sum_{n=1}^{\infty} \frac{n^2}{2^n}$
$$
\begin{array}{lll}
\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}&=\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left|\frac{n^2}{2^n}\right|}&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\frac{n^2}{2^n}}&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{\sqrt[n]{n^2}}{\sqrt[n]{2^n}}&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{(n^{1/n})^2}{2}&\mbox{}\\
&=\displaystyle \frac{1^2}{2}&\mbox{}\\
&=\displaystyle \frac{1}{2}&\mbox{}\\
&\lt 1&\mbox{}\\
\end{array}
$$
Thus, by the Root Test, the series converges absolutely.
$\displaystyle \sum_{n=1}^{\infty} \frac{2^n}{n^3}$
$$
\begin{array}{lll}
\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}&=\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left|\frac{2^n}{n^3}\right|}&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\frac{2^n}{n^3}}&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{\sqrt[n]{2^n}}{\sqrt[n]{n^3}}&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{2}{(n^{1/n})^3}&\mbox{}\\
&=\displaystyle \frac{2}{1^3}&\mbox{}\\
&=\displaystyle 2&\mbox{}\\
&\gt \displaystyle 1&\mbox{}\\
\end{array}
$$
Thus, by the Root Test, the series diverges.
$\displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{1+n}\right)^n$
$$
\begin{array}{lll}
\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}&=\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left|\left(\frac{1}{1+n}\right)^n\right|}&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left(\frac{1}{1+n}\right)^n}&\mbox{}\\
&=\displaystyle \lim_{n \rightarrow \infty} \frac{1}{1+n}&\mbox{}\\
&=\displaystyle 0&\mbox{}\\
&\lt \displaystyle 1&\mbox{}\\
\end{array}
$$
Thus, by the Root Test, the series converges absolutely.
Pearl of Wisdom #2
The Root Test is often used for series where $|a_n|=b_n^n.$
Dire Warning #2
If $\displaystyle \sum_{n=1}^{\infty} a_n$ is a series of nonzero terms an $$\rho=\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}.$$
The sum of the series $\displaystyle \sum_{n=1}^{\infty} a_n$ is NOT $\rho.$
Bonus Example
Let $\displaystyle a_n=\begin{cases} \displaystyle \frac{n}{2^n} & \mbox{ if $n$ is odd} \\ \displaystyle \frac{1}{2^n} & \mbox{ if $n$ is even} \\\end{cases}$
Does the series $\displaystyle \sum_{n=1}^{\infty} a_n$ converge?
We shall attempt to apply the Ratio Test.
Case 1: Suppose $n$ is odd. Then $n+1$ is even so that $$ \begin{array}{lll} \displaystyle \left|\frac{a_{n+1}}{a_n}\right|&=\displaystyle \left|\frac{\frac{1}{2^{n+1}}}{\frac{n}{2^n}}\right|&\mbox{}\\ &=\displaystyle \left|\frac{\frac{1}{2^{n+1}}}{\frac{n}{2^n}}\right|&\mbox{}\\ &=\displaystyle \frac{\frac{1}{2^{n+1}}}{\frac{n}{2^n}}&\mbox{}\\ &=\displaystyle \frac{1}{2^{n+1}}\cdot\frac{2^n}{n}&\mbox{}\\ &=\displaystyle \frac{1}{n}\cdot\frac{2^n}{2^{n+1}}&\mbox{}\\ &=\displaystyle \frac{1}{n}\cdot\frac{1}{2}&\mbox{}\\ &=\displaystyle \frac{1}{2n}&\mbox{}\\ \end{array} $$ Thus, the subsequence of ratios where $n$ is odd approaches $0$ as $n \rightarrow \infty.$
Case 2: Suppose $n$ is even. Then $n+1$ is odd so that $$ \begin{array}{lll} \displaystyle \left|\frac{a_{n+1}}{a_n}\right|&=\displaystyle \left|\frac{\frac{n+1}{2^{n+1}}}{\frac{1}{2^n}}\right|&\mbox{}\\ &=\displaystyle \frac{\frac{n+1}{2^{n+1}}}{\frac{1}{2^n}}&\mbox{}\\ &=\displaystyle \frac{n+1}{2^{n+1}}\cdot\frac{2^n}{1}&\mbox{}\\ &=\displaystyle \frac{n+1}{2}&\mbox{}\\ \end{array} $$ Thus, the subsequence of ratios where $n$ is even approaches $\infty$ as $n \rightarrow \infty.$
We conclude that $\displaystyle \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|$ does not exist.
Therefore, the Ratio Test does not apply.
We shall now attempt the Root Test using the same strategy as above.
Case 1: Suppose $n$ is odd. Then $$ \begin{array}{lll} \displaystyle \sqrt[n]{\left|a_n\right|}&=\displaystyle \sqrt[n]{\left|\frac{n}{2^n}\right|}&\mbox{}\\ &=\displaystyle \sqrt[n]{\frac{n}{2^n}}&\mbox{}\\ &=\displaystyle \frac{\sqrt[n]{n}}{\sqrt[n]{2^n}}&\mbox{}\\ &=\displaystyle \frac{n^{1/n}}{2}&\mbox{}\\ \end{array} $$ Thus, the subsequence of $n$th roots where $n$ is odd approaches $\displaystyle \frac{1}{2}$ as $n \rightarrow \infty.$
Case 2: Suppose $n$ is even. Then $$ \begin{array}{lll} \displaystyle \sqrt[n]{\left|a_n\right|}&=\displaystyle \sqrt[n]{\left|\frac{1}{2^n}\right|}&\mbox{}\\ &=\displaystyle \sqrt[n]{\frac{1}{2^n}}&\mbox{}\\ &=\displaystyle \frac{1}{2}&\mbox{}\\ \end{array} $$ Thus, the subsequence of $n$th roots where $n$ is even also approaches $\displaystyle \frac{1}{2}$ as $n \rightarrow \infty.$
Since both subsequences converge to $\displaystyle \frac{1}{2},$ we have that $\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}=\frac{1}{2}\lt 1.$
By the Root Test, the series converges absolutely.
Case 1: Suppose $n$ is odd. Then $n+1$ is even so that $$ \begin{array}{lll} \displaystyle \left|\frac{a_{n+1}}{a_n}\right|&=\displaystyle \left|\frac{\frac{1}{2^{n+1}}}{\frac{n}{2^n}}\right|&\mbox{}\\ &=\displaystyle \left|\frac{\frac{1}{2^{n+1}}}{\frac{n}{2^n}}\right|&\mbox{}\\ &=\displaystyle \frac{\frac{1}{2^{n+1}}}{\frac{n}{2^n}}&\mbox{}\\ &=\displaystyle \frac{1}{2^{n+1}}\cdot\frac{2^n}{n}&\mbox{}\\ &=\displaystyle \frac{1}{n}\cdot\frac{2^n}{2^{n+1}}&\mbox{}\\ &=\displaystyle \frac{1}{n}\cdot\frac{1}{2}&\mbox{}\\ &=\displaystyle \frac{1}{2n}&\mbox{}\\ \end{array} $$ Thus, the subsequence of ratios where $n$ is odd approaches $0$ as $n \rightarrow \infty.$
Case 2: Suppose $n$ is even. Then $n+1$ is odd so that $$ \begin{array}{lll} \displaystyle \left|\frac{a_{n+1}}{a_n}\right|&=\displaystyle \left|\frac{\frac{n+1}{2^{n+1}}}{\frac{1}{2^n}}\right|&\mbox{}\\ &=\displaystyle \frac{\frac{n+1}{2^{n+1}}}{\frac{1}{2^n}}&\mbox{}\\ &=\displaystyle \frac{n+1}{2^{n+1}}\cdot\frac{2^n}{1}&\mbox{}\\ &=\displaystyle \frac{n+1}{2}&\mbox{}\\ \end{array} $$ Thus, the subsequence of ratios where $n$ is even approaches $\infty$ as $n \rightarrow \infty.$
We conclude that $\displaystyle \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|$ does not exist.
Therefore, the Ratio Test does not apply.
We shall now attempt the Root Test using the same strategy as above.
Case 1: Suppose $n$ is odd. Then $$ \begin{array}{lll} \displaystyle \sqrt[n]{\left|a_n\right|}&=\displaystyle \sqrt[n]{\left|\frac{n}{2^n}\right|}&\mbox{}\\ &=\displaystyle \sqrt[n]{\frac{n}{2^n}}&\mbox{}\\ &=\displaystyle \frac{\sqrt[n]{n}}{\sqrt[n]{2^n}}&\mbox{}\\ &=\displaystyle \frac{n^{1/n}}{2}&\mbox{}\\ \end{array} $$ Thus, the subsequence of $n$th roots where $n$ is odd approaches $\displaystyle \frac{1}{2}$ as $n \rightarrow \infty.$
Case 2: Suppose $n$ is even. Then $$ \begin{array}{lll} \displaystyle \sqrt[n]{\left|a_n\right|}&=\displaystyle \sqrt[n]{\left|\frac{1}{2^n}\right|}&\mbox{}\\ &=\displaystyle \sqrt[n]{\frac{1}{2^n}}&\mbox{}\\ &=\displaystyle \frac{1}{2}&\mbox{}\\ \end{array} $$ Thus, the subsequence of $n$th roots where $n$ is even also approaches $\displaystyle \frac{1}{2}$ as $n \rightarrow \infty.$
Since both subsequences converge to $\displaystyle \frac{1}{2},$ we have that $\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}=\frac{1}{2}\lt 1.$
By the Root Test, the series converges absolutely.