Learning Outcomes
- Use the ratio and root tests to determine if a series converges or diverges.
Section 16.7 Ratio and Root Tests (SQ7)
Subsection 16.7.1 Activities
Activity 16.100.
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 16.101.
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 16.102.
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 16.103.
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 16.104.
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 16.105. 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 16.106. 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 16.107.
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 16.108.
(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 16.109.
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 16.110.
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 16.111.
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 16.112.
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 16.7.2 Videos
Subsection 16.7.3 Exercises
Exercises available at
https://tbil.org/preview/calculus/exercises/#/bank/SQ7/
