Ratio Test & Root Test

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Ratio Test

Given the series $\sum_{n=1}^\infty a_n$

1. If $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=L<1$, then the series is absolutely convergent.
2. If $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=L>1$ or $\operatorname*{lim}_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\infty$, then the series is divergent.
3. If $\operatorname*{lim}_{n\to\infty}\left|{\frac{a_{n+1}}{a_{n}}}\right|=L=1$, then the series might converge or diverge. No conclusion can be made, and use another test.

Example

Determine if the series $\sum_{n=1}^\infty\frac{n3^n}{4^{n-1}}$ converges absolutely.

Solution

Let $a_n=\frac{n3^n}{4^{n-1}}$. Then, $a_{n+1}=\frac{(n+1)3^{n+1}}{4^{(n+1)-1}}=\frac{(n+1)3^{n+1}}{4^{n}}.$

$\begin{gathered} \operatorname*{lim}_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| =\operatorname*{lim}_{n\to\infty}\left|\frac{\frac{(n+1)3^{n+1}}{4^{n}}}{\frac{n3^{n}}{4^{n-1}}}\right|=\operatorname*{lim}_{n\to\infty}\left|\left(\frac{(n+1)3^{n+1}}{4^{n}}\right)\left(\frac{4^{n-1}}{n3^{n}}\right)\right| \\ =\lim_{n\to\infty}\left|\frac{3(n+1)}{4n}\right|=\lim_{n\to\infty}\left|\frac{3(1+1/n)}4\right|=\frac34<1  \end{gathered}$

By the Ratio Test, $\sum_{n=1}^\infty\frac{n3^n}{4^{n-1}}$ is absolutely convergent.

Root Test

Given the series $\sum_{n=1}^\infty a_n$.

1. If $\operatorname*{lim}_{n\to\infty}\sqrt[n]{\left|a_{n}\right|}=L<1$, then the series is absolutely convergent.
2. If $\operatorname*{lim}_{n\to\infty}\sqrt[n]{\left|a_{n}\right|}=L>1$ or $\operatorname*{lim}_{n\to\infty}\sqrt[n]{\left|a_{n}\right|}= \infty$, then the series is divergent.
3. If $\operatorname*{lim}_{n\to\infty}\sqrt[n]{\left|a_{n}\right|}=L=1$, then the series might converge or diverge. No conclusion can be made, and use another test (do not use the Ratio Test because $L$ will be 1 again.)

Example

Determine if the series $\sum_{n=1}^\infty\frac{n^n}{3^{1+3n}}$ absolutely converges.

Solution

Let $a_{n}=\frac{n^{n}}{3^{1+3n}}$.

$$\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\lim_{n\to\infty}\sqrt[n]{\left|\frac{n^n}{3^{1+3n}}\right|}=\lim_{n\to\infty}\left(\frac{n^n}{3^{1+3n}}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\frac{n}{3^{1/n+3}}=\infty$$

By the Root Test, $\sum_{n=1}^\infty\frac{n^n}{3^{1+3n}}$ diverges.