Learning Outcomes
- Use the ratio and root tests to determine if a series converges or diverges.
Section 8.7 Ratio and Root Tests (SQ7)
Subsection 8.7.1 Activities
Activity 8.7.1.
Consider the series \(\displaystyle \sum_{n=0}^\infty \frac{2^n}{3^n-2}.\)
(a)
Which of these series most closely resembles \(\displaystyle \sum_{n=0}^\infty \frac{2^n}{3^n-2}\text{?}\)
- \(\displaystyle \sum_{n=0}^\infty \frac{2}{3}\text{.}\)
- \(\displaystyle \sum_{n=0}^\infty \frac{2}{3}n\text{.}\)
- \(\displaystyle \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n\text{.}\)
(b)
Based on your previous choice, do we think this series is more likely to converge or diverge?
(c)
Find \(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=\lim_{n\to\infty}\frac{2^{n+1}(3^n-2)}{(3^{n+1}-2)2^n}=\lim_{n\to\infty}\frac{2\cdot 2^{n}(3^n-2)}{3(3^{n}-\frac{2}{3})2^n}.\)
- \(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=0\text{.}\)
- \(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=\frac{2}{3}\text{.}\)
- \(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=1\text{.}\)
- \(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=2\text{.}\)
- \(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=3\text{.}\)
Activity 8.7.2.
Consider the series \(\displaystyle \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty \frac{3}{2^n}\text{.}\)
(a)
Does \(\displaystyle \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty \frac{3}{2^n}\) converge?
(b)
Find \(\displaystyle\frac{a_{n+1}}{a_n}\text{.}\)
- \(2\text{.}\)
- \(\displaystyle \frac{1}{2}\text{.}\)
- \(\displaystyle \frac{2^n}{2^n+1}\text{.}\)
- \(\displaystyle \frac{9}{2^{2n+1}}\text{.}\)
- \(\displaystyle \frac{9}{2^{n+2}}\text{.}\)
(c)
Find \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\)
- \(-\infty\text{.}\)
- \(0\text{.}\)
- \(\displaystyle \frac{1}{2}\text{.}\)
- \(2\text{.}\)
- \(\infty\text{.}\)
Activity 8.7.3.
Consider the series \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{n^2}{n+1}\text{.}\)
(a)
Does \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{n^2}{n+1}\) converge?
Answer.
No.
(b)
Find \(\displaystyle\frac{a_{n+1}}{a_n}\text{.}\)
- \(\displaystyle 1+\frac{n+1}{n^2}\text{.}\)
- \(\displaystyle \frac{(n^2+1)(n+1)}{(n+2)n^2}\text{.}\)
- \(\displaystyle \frac{(n+1)}{(n+2)n^2}\text{.}\)
- \(\displaystyle \frac{(n+1)^3}{(n+2)n^2}\text{.}\)
- \(\displaystyle \frac{(n+1)n^2}{n+2}\text{.}\)
Answer.
D. \(\displaystyle \frac{(n+1)^3}{(n+2)n^2}\text{.}\)
A. results from bad subscript usage. B. occurs from using the Freshman Dream Theorem: \((n+1)^2=n^2+1\text{.}\) C. results from a bad cancellation. E. is the reciprocal of C.
(c)
Find \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\)
- \(1\text{.}\)
- \(0\text{.}\)
- \(\displaystyle \frac{1}{2}\text{.}\)
- \(2\text{.}\)
- \(\infty\text{.}\)
Answer.
A. \(1\text{.}\)
Activity 8.7.4.
Consider the series \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{1}{n}\text{.}\)
(a)
Does \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{1}{n}\) converge?
(b)
Find \(\displaystyle\frac{a_{n+1}}{a_n}\text{.}\)
(c)
Find \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\)
Activity 8.7.5.
Consider the series \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{(-1)^n}{n}\text{.}\)
(a)
Does \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{(-1)^n}{n}\) converge?
(b)
Find \(\frac{a_{n+1}}{a_n}\text{.}\)
(c)
Find \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\)
Fact 8.7.6. The Ratio Test.
Let \(\displaystyle\sum a_n\) be a series and suppose that \(\displaystyle\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\rho\text{.}\) Then
- \(\displaystyle\sum a_n\) converges if \(\rho\) is less than 1, and
- \(\displaystyle\sum a_n\) diverges if \(\rho\) is greater than 1.
- If \(\rho=1\text{,}\) we cannot determine if \(\displaystyle\sum a_n\) converges or diverges with this method.
Fact 8.7.7. The Root Test.
Let \(N\) be an integer and let \(\displaystyle\sum a_n\) be a series with \(a_n\geq 0\) for \(n\geq N\text{,}\) and suppose that \(\displaystyle\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|}=\rho\text{.}\) Then
- \(\displaystyle\sum a_n\) converges if \(\rho\) is less than 1, and
- \(\displaystyle\sum a_n\) diverges if \(\rho\) is greater than 1.
- If \(\rho=1\text{,}\) we cannot determine if \(\displaystyle\sum a_n\) converges or diverges with this method.
Activity 8.7.8.
Consider the series \(\displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}\text{.}\)
(a)
Which of the following is \(a_n\text{?}\)
- \(n^2\text{.}\)
- \(n!\text{.}\)
- \(\displaystyle\frac{n^2}{n!}\text{.}\)
(b)
Which of the following is \(a_{n+1}\text{?}\)
- \(\displaystyle\frac{n^2}{n!}\text{.}\)
- \(\displaystyle(n+1)^2\text{.}\)
- \(\displaystyle(n+1)!\text{.}\)
- \(\displaystyle\frac{(n+1)^2}{(n+1)!}\text{.}\)
- \(\displaystyle\frac{n^2+1}{n!+1}\text{.}\)
(c)
Which of the following is \(\displaystyle\left|\frac{a_{n+1}}{a_n}\right|\text{?}\)
- \(\displaystyle\frac{(n+1)^2n^2}{(n+1)!n!}\text{.}\)
- \(\displaystyle\frac{(n+1)^2n!}{(n+1)!n^2}\text{.}\)
- \(\displaystyle\frac{(n+1)!n!}{(n+1)^2n^2}\text{.}\)
- \(\displaystyle\frac{(n+1)!n^2}{(n+1)^2n!}\text{.}\)
(d)
Using the fact \((n+1)!=(n+1)\cdot n!\text{,}\) simplify \(\displaystyle\left|\frac{a_{n+1}}{a_n}\right|\) as much as possible.
(e)
Find \(\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\text{.}\)
(f)
Does \(\displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}\) converge?
Activity 8.7.9.
(a)
What is \(a_n\text{?}\)
(b)
Which of the following is \(\displaystyle\sqrt[n]{|a_n|}\text{?}\)
- \(\displaystyle \frac{n+1}{9}\text{.}\)
- \(\displaystyle \frac{n}{9}\text{.}\)
- \(n\text{.}\)
- \(9\text{.}\)
- \(\displaystyle \frac{1}{9}\text{.}\)
(c)
Find \(\displaystyle\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|}\text{.}\)
(d)
Does \(\displaystyle \sum_{n=1}^\infty \frac{n^n}{9^n}\) converge?
Activity 8.7.10.
For each series, use the ratio or root test to determine if the series converges or diverges.
(a)
\(\displaystyle \sum_{n=1}^\infty \displaystyle\left(\frac{1}{1+n}\right)^n\)
(b)
\(\displaystyle \sum_{n=1}^\infty \displaystyle\frac{2^n}{n^n}\)
(c)
\(\displaystyle \sum_{n=1}^\infty \displaystyle\frac{(2n)!}{(n!)(n!)}\)
(d)
\(\displaystyle \sum_{n=1}^\infty \displaystyle\frac{4^n(n!)(n!)}{(2n)!}\)
Activity 8.7.11.
Consider the series \(\displaystyle \sum_{n=0}^\infty \displaystyle\frac{2^n+5}{3^n}\text{.}\)
(a)
Use the root test to check for convergence of this series.
(b)
Use the ratio test to check for convergence of this series.
(c)
Use the comparison (or limit comparison) test to check for convergence of this series.
(d)
Find the sum of this series.
Activity 8.7.12.
Consider \(\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}\text{.}\) Recall that \(\displaystyle \sqrt[n]{\frac{n}{3^n}}=\left(\frac{n}{3^n}\right)^{1/n}=\frac{n^{1/n}}{(3^n)^{1/n}}.\)
(a)
Let \(\displaystyle \alpha=\lim_{n\to\infty}\ln(n^{1/n})=\lim_{n\to\infty}\frac{1}{n}\ln(n)\text{.}\) Find \(\alpha\text{.}\)
(b)
Recall that \(\displaystyle \lim_{n\to\infty}n^{1/n}=\lim_{n\to\infty} e^{\ln(n^{1/n})}=e^\alpha.\) Find \(\displaystyle \lim_{n\to\infty}n^{1/n}\text{.}\)
(c)
Find \(\displaystyle \lim_{n\to\infty} \sqrt[n]{\frac{n}{3^n}}=\lim_{n\to\infty}\left(\frac{n}{3^n}\right)^{1/n}=\lim_{n\to\infty}\frac{n^{1/n}}{(3^n)^{1/n}}\text{.}\)
(d)
Does \(\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}\) converge?
Activity 8.7.13.
Consider the series \(\displaystyle \sum_{n=0}^\infty \displaystyle\frac{n^2}{2^n}\text{.}\)
(a)
Use the root test to check for convergence of this series.
(b)
Use the ratio test to check for convergence of this series.
(c)
Use the comparison (or limit comparison) test to check for convergence of this series.
Subsection 8.7.2 Videos
Subsection 8.7.3 Exercises
Exercises available at
https://tbil.org/calculus/preview/exercises/#/bank/SQ7/
