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Calculus 2 : Ratio Test

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Example Questions

Example Question #62 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{11^n}{6^{3n}}\newline note: \hspace f(n) = \frac{11^n}{6^{3n}}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{216}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{3}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{216}$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{11^{\left(n+1\right)}}{6^{3\left(n+1\right)}}}{\frac{11^n}{6^{3n}}} = \frac{6^{3n}\cdot \:11^{n+1}}{6^{3\left(n+1\right)}\cdot \:11^n} = \frac{11}{216}$

$\lim_{n\rightarrow \infty} \frac{11}{216} = \frac{11}{216}$

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Example Question #31 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{10\cdot \:7^n}{2^{3n}}\newline note: \hspace f(n) = \frac{10\cdot \:7^n}{2^{3n}}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{8}{7}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{8}$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{8}$

Explanation:

$\frac{\frac{7^{n+1}}{2^{3\left(n+1\right)}}}{\frac{7^n}{2^{3n}}} = \frac{2^{3n}\cdot \:7^{n+1}}{2^{3\left(n+1\right)}\cdot \:7^n} = \frac{7}{8}$

$\lim_{n\rightarrow \infty}\frac{7}{8} =\frac{7}{8}$

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Example Question #32 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{11\cdot \:20^n}{4^{2n}}\newline note: \hspace f(n) = \frac{11\cdot \:20^n}{4^{2n}}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{4}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{20}$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{4}$

Explanation:

$\frac{\frac{20^{\left(n+1\right)}}{4^{2\left(n+1\right)}}}{\frac{20^n}{4^{2n}}} = \frac{4^{2n}\cdot \:20^{n+1}}{4^{2\left(n+1\right)}\cdot \:20^n} = \frac{20}{16} = \frac{5}{4}$

$\lim_{n\rightarrow \infty} \frac{5}{4} = \frac{5}{4}$

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Example Question #2851 : Calculus Ii

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{n7^n}{5^{2n}}\newline note: \hspace f(n) = \frac{n7^n}{5^{2n}}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{25}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{25}{7}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{25}$

Explanation:

$\frac{\frac{7^{n+1}\left(n+1\right)}{5^{2\left(n+1\right)}}}{\frac{7^nn}{5^{2n}}} = \frac{5^{2n}\cdot \:7^{n+1}\left(n+1\right)}{5^{2\left(n+1\right)}\cdot \:7^nn} = \frac{7\left(n+1\right)}{25n}$

$\lim_{n\rightarrow \infty} \frac{7\left(n+1\right)}{25n} = \frac{7}{25}$

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Example Question #35 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{7^n}{n5^{2n}}\newline note: \hspace f(n) = \frac{7^n}{n5^{2n}}$

Possible Answers:

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{25}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{25}{7}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{25}$

Explanation:

$\frac{\frac{7^{n+1}}{5^{2\left(n+1\right)}\left(n+1\right)}}{\frac{7^n}{5^{2n}n}}= \frac{5^{2n}\cdot \:7^{n+1}n}{5^{2\left(n+1\right)}\cdot \:7^n\left(n+1\right)} = \frac{7n}{25\left(n+1\right)}$

$\lim_{n\rightarrow \infty} \frac{7n}{25\left(n+1\right)} = \frac{7}{25}$

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Example Question #36 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{n4^n}{3^{2n}}\newline note: \hspace f(n) = \frac{n4^n}{3^{2n}}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{9}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{9}{4}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{9}$

Explanation:

$\frac{\frac{4^{n+1}\left(n+1\right)}{3^{2\left(n+1\right)}}}{\frac{4^nn}{3^{2n}}} = \frac{3^{2n}\cdot \:4^{n+1}\left(n+1\right)}{3^{2\left(n+1\right)}\cdot \:4^nn} = \frac{4\left(n+1\right)}{9n}$

$\lim_{n\rightarrow \infty} \frac{4\left(n+1\right)}{9n} = \frac{4}{9}$

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Example Question #31 : Convergence And Divergence

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{11^n}{n5^{3n}}\newline note: \hspace f(n) = \frac{11^n}{n5^{3n}}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{125}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{125}{11}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{125}$

Explanation:

$\frac{\frac{11^{n+1}}{5^{3\left(n+1\right)}\left(n+1\right)}}{\frac{11^n}{5^{3n}n}} = \frac{5^{3n}\cdot \:11^{n+1}n}{5^{3\left(n+1\right)}\cdot \:11^n\left(n+1\right)} = \frac{11n}{125\left(n+1\right)}$

$\lim_{n\rightarrow \infty} \frac{11n}{125\left(n+1\right)} = \frac{11}{125}$

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Example Question #38 : Ratio Test

$\\ \textup{Use the ratio test to find out if the following series is convergent:}\\ \sum_{n = 1}^{\infty} \frac{9^n}{4^{2n}}\newline note: \hspace f(n) = \frac{9^n}{4^{2n}}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{9}{16}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{16}{9}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{9}{16}$

Explanation:

$\frac{\frac{9^{\left(n+1\right)}}{4^{2\left(n+1\right)}}}{\frac{9^n}{4^{2n}}} = \frac{4^{2n}\cdot \:9^{n+1}}{4^{2\left(n+1\right)}\cdot \:9^n} = \frac{9}{16}$

$\lim_{n\rightarrow \infty} \frac{9}{16} = \frac{9}{16}$

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Example Question #2851 : Calculus Ii

$\\ \textup{Use the ratio test to find out if the following series is convergent:}\\ \sum_{n = 1}^{\infty} \frac{4^{2n}}{5\cdot \:6^{2n}}\newline note: \hspace f(n) = \frac{4^{2n}}{5\cdot \:6^{2n}}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{9}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 9$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{9}$

Explanation:

$\frac{\left(\frac{2}{3}\right)^{2\left(n+1\right)}}{\left(\frac{2}{3}\right)^{2n}}=\left(\frac{2}{3}\right)^2=\frac{4}{9}$

$\lim_{n\rightarrow \infty} \frac{4}{9} = \frac{4}{9}$

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Example Question #40 : Ratio Test

$\\ \textup{Use the ratio test to find out if the following series is convergent:}\\ \sum_{n = 1}^{\infty} \frac{7^n}{5^{2n}}\newline note: \hspace f(n) = \frac{7^n}{5^{2n}}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{25}{7}$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{25}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{25}$

Explanation:

$\frac{\frac{7^{n+1}}{5^{2\left(n+1\right)}}}{\frac{7^n}{5^{2n}}} = \frac{5^{2n}\cdot \:7^{n+1}}{5^{2\left(n+1\right)}\cdot \:7^n} = \frac{7}{25}$

$\lim_{n\rightarrow \infty} \frac{7}{25} = \frac{7}{25}$

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Calculus 2 : Ratio Test

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #62 : Series In Calculus

Example Question #31 : Ratio Test

Example Question #32 : Ratio Test

Example Question #2851 : Calculus Ii

Example Question #35 : Ratio Test

Example Question #36 : Ratio Test

Example Question #31 : Convergence And Divergence

Example Question #38 : Ratio Test

Example Question #2851 : Calculus Ii

Example Question #40 : Ratio Test

All Calculus 2 Resources

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