Ratio Test
Given the series \(\sum_{n=1}^\infty a_n\)
- If \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=L<1\), then the series is absolutely convergent.
- 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.
- 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 \(\sum_{n=1}^\infty a_n\).
- If \(\operatorname*{lim}_{n\to\infty}\sqrt[n]{\left|a_{n}\right|}=L<1\), then the series is absolutely convergent.
- 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.
- 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.