TBIL-Cal Ratio and Root Tests (SQ7)

Section 8.7 Ratio and Root Tests (SQ7)

Learning Outcomes

Use the ratio and root tests to determine if a series converges or diverges.
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Subsection 8.7.1 Activities

Activity 8.7.1.

Consider the series \(\displaystyle \sum_{n=0}^\infty \frac{2^n}{3^n-2}.\)

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(a)

Which of these series most closely resembles \(\displaystyle \sum_{n=0}^\infty \frac{2^n}{3^n-2}\text{?}\)

\(\displaystyle \sum_{n=0}^\infty \frac{2}{3}\text{.}\)

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\(\displaystyle \sum_{n=0}^\infty \frac{2}{3}n\text{.}\)

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\(\displaystyle \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n\text{.}\)

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(b)

Based on your previous choice, do we think this series is more likely to converge or diverge?

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(c)

Find \(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=\lim_{n\to\infty}\frac{2^{n+1}(3^n-2)}{(3^{n+1}-2)2^n}=\lim_{n\to\infty}\frac{2\cdot 2^{n}(3^n-2)}{3(3^{n}-\frac{2}{3})2^n}.\)

\(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=0\text{.}\)

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\(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=\frac{2}{3}\text{.}\)

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\(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=1\text{.}\)

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\(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=2\text{.}\)

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\(\displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=3\text{.}\)

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Activity 8.7.2.

Consider the series \(\displaystyle \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty \frac{3}{2^n}\text{.}\)

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(a)

Does \(\displaystyle \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty \frac{3}{2^n}\) converge?

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(b)

Find \(\displaystyle\frac{a_{n+1}}{a_n}\text{.}\)

\(2\text{.}\)

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\(\displaystyle \frac{1}{2}\text{.}\)

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\(\displaystyle \frac{2^n}{2^n+1}\text{.}\)

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\(\displaystyle \frac{9}{2^{2n+1}}\text{.}\)

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\(\displaystyle \frac{9}{2^{n+2}}\text{.}\)

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(c)

Find \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\)

\(-\infty\text{.}\)

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\(0\text{.}\)

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\(\displaystyle \frac{1}{2}\text{.}\)

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\(2\text{.}\)

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\(\infty\text{.}\)

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Activity 8.7.3.

Consider the series \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{n^2}{n+1}\text{.}\)

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(a)

Does \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{n^2}{n+1}\) converge?

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(b)

Find \(\displaystyle\frac{a_{n+1}}{a_n}\text{.}\)

\(\displaystyle 1+\frac{n+1}{n^2}\text{.}\)

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\(\displaystyle \frac{(n^2+1)(n+1)}{(n+2)n^2}\text{.}\)

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\(\displaystyle \frac{(n+1)}{(n+2)n^2}\text{.}\)

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\(\displaystyle \frac{(n+1)^3}{(n+2)n^2}\text{.}\)

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\(\displaystyle \frac{(n+1)n^2}{n+2}\text{.}\)

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(c)

Find \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\)

\(1\text{.}\)

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\(0\text{.}\)

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\(\displaystyle \frac{1}{2}\text{.}\)

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\(2\text{.}\)

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\(\infty\text{.}\)

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Activity 8.7.4.

Consider the series \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{1}{n}\text{.}\)

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(a)

Does \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{1}{n}\) converge?

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(b)

Find \(\displaystyle\frac{a_{n+1}}{a_n}\text{.}\)

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(c)

Find \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\)

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Activity 8.7.5.

Consider the series \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{(-1)^n}{n}\text{.}\)

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(a)

Does \(\displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{(-1)^n}{n}\) converge?

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(b)

Find \(\frac{a_{n+1}}{a_n}\text{.}\)

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(c)

Find \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\)

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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

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\(\displaystyle\sum a_n\) diverges if \(\rho\) is greater than 1.

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If \(\rho=1\text{,}\) we cannot determine if \(\displaystyle\sum a_n\) converges or diverges with this method.

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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

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\(\displaystyle\sum a_n\) diverges if \(\rho\) is greater than 1.

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If \(\rho=1\text{,}\) we cannot determine if \(\displaystyle\sum a_n\) converges or diverges with this method.

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Activity 8.7.8.

Consider the series \(\displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}\text{.}\)

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(a)

Which of the following is \(a_n\text{?}\)

\(n^2\text{.}\)

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\(n!\text{.}\)

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\(\displaystyle\frac{n^2}{n!}\text{.}\)

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(b)

Which of the following is \(a_{n+1}\text{?}\)

\(\displaystyle\frac{n^2}{n!}\text{.}\)

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\(\displaystyle(n+1)^2\text{.}\)

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\(\displaystyle(n+1)!\text{.}\)

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\(\displaystyle\frac{(n+1)^2}{(n+1)!}\text{.}\)

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\(\displaystyle\frac{n^2+1}{n!+1}\text{.}\)

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(c)

Which of the following is \(\displaystyle\left|\frac{a_{n+1}}{a_n}\right|\text{?}\)

\(\displaystyle\frac{(n+1)^2n^2}{(n+1)!n!}\text{.}\)

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\(\displaystyle\frac{(n+1)^2n!}{(n+1)!n^2}\text{.}\)

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\(\displaystyle\frac{(n+1)!n!}{(n+1)^2n^2}\text{.}\)

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\(\displaystyle\frac{(n+1)!n^2}{(n+1)^2n!}\text{.}\)

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(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.

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(e)

Find \(\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\text{.}\)

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(f)

Does \(\displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}\) converge?

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Activity 8.7.9.

(a)

What is \(a_n\text{?}\)

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(b)

Which of the following is \(\displaystyle\sqrt[n]{|a_n|}\text{?}\)

\(\displaystyle \frac{n+1}{9}\text{.}\)

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\(\displaystyle \frac{n}{9}\text{.}\)

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\(n\text{.}\)

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\(9\text{.}\)

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\(\displaystyle \frac{1}{9}\text{.}\)

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(c)

Find \(\displaystyle\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|}\text{.}\)

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(d)

Does \(\displaystyle \sum_{n=1}^\infty \frac{n^n}{9^n}\) converge?

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Activity 8.7.10.

For each series, use the ratio or root test to determine if the series converges or diverges.

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(a)

\(\displaystyle \sum_{n=1}^\infty \displaystyle\left(\frac{1}{1+n}\right)^n\)

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(b)

\(\displaystyle \sum_{n=1}^\infty \displaystyle\frac{2^n}{n^n}\)

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(c)

\(\displaystyle \sum_{n=1}^\infty \displaystyle\frac{(2n)!}{(n!)(n!)}\)

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(d)

\(\displaystyle \sum_{n=1}^\infty \displaystyle\frac{4^n(n!)(n!)}{(2n)!}\)

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Activity 8.7.11.

Consider the series \(\displaystyle \sum_{n=0}^\infty \displaystyle\frac{2^n+5}{3^n}\text{.}\)

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(a)

Use the root test to check for convergence of this series.

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(b)

Use the ratio test to check for convergence of this series.

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(c)

Use the comparison (or limit comparison) test to check for convergence of this series.

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(d)

Find the sum of this series.

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Activity 8.7.12.

Consider \(\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}\text{.}\) Recall that \(\displaystyle \sqrt[n]{\frac{n}{3^n}}=\left(\frac{n}{3^n}\right)^{1/n}=\frac{n^{1/n}}{(3^n)^{1/n}}.\)

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(a)

Let \(\displaystyle \alpha=\lim_{n\to\infty}\ln(n^{1/n})=\lim_{n\to\infty}\frac{1}{n}\ln(n)\text{.}\) Find \(\alpha\text{.}\)

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(b)

Recall that \(\displaystyle \lim_{n\to\infty}n^{1/n}=\lim_{n\to\infty} e^{\ln(n^{1/n})}=e^\alpha.\) Find \(\displaystyle \lim_{n\to\infty}n^{1/n}\text{.}\)

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(c)

Find \(\displaystyle \lim_{n\to\infty} \sqrt[n]{\frac{n}{3^n}}=\lim_{n\to\infty}\left(\frac{n}{3^n}\right)^{1/n}=\lim_{n\to\infty}\frac{n^{1/n}}{(3^n)^{1/n}}\text{.}\)

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(d)

Does \(\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}\) converge?

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Activity 8.7.13.

Consider the series \(\displaystyle \sum_{n=0}^\infty \displaystyle\frac{n^2}{2^n}\text{.}\)

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(a)

Use the root test to check for convergence of this series.

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(b)

Use the ratio test to check for convergence of this series.