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Calculus 2 : Ratio Test

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Example Questions

Example Question #2861 : Calculus Ii

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

Possible Answers:

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

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

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

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{10}{9}$

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

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{10}{9}$

Explanation:

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

$\frac{\frac{10^{\left(n+1\right)}}{3^{2\left(n+1\right)}}}{\frac{10^n}{3^{2n}}} = \frac{3^{2n}\cdot \:10^{n+1}}{3^{2\left(n+1\right)}\cdot \:10^n} = \frac{10}{9}$

$\lim_{n\rightarrow \infty} \frac{10}{9} = \frac{10}{9}$

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Example Question #2861 : Calculus Ii

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{50n!}{n^{4}ln(n)} \newline note: \hspace f(n) = \frac{50n!}{n^{4}ln(n)}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

$\lim_{n\rightarrow \infty} \frac{n^4ln(n)}{(n+1)^3ln(n+1)} = \infty$

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Example Question #2861 : Calculus Ii

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{70n!}{n^{8}ln(n)}\newline note: \hspace f(n) = \frac{70n!}{n^{8}ln(n)}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$\frac{\frac{\left(n+1\right)!}{\left(n+1\right)^8\ln \left(\left(n+1\right)\right)}}{\frac{n!}{n^8\ln \left(n\right)}} = \frac{(n+1)!n^8ln(n)}{n!(n+1)^8ln(n+1)} = \frac{n^8ln(n)}{(n+1)^7ln(n+1)}$

$\lim_{n\rightarrow \infty} \frac{n^8ln(n)}{(n+1)^7ln(n+1)} = \infty$

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Example Question #41 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{35n!}{17n^{10}ln(n)}\newline note: \hspace f(n) = \frac{35n!}{17n^{10}ln(n)}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{10}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 10$

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

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

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

Correct answer:

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

Explanation:

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

$\frac{\frac{\left(n+1\right)!}{\left(n+1\right)^{10}\ln \left(\left(n+1\right)\right)}}{\frac{n!}{n^{10}\ln \left(n\right)}} = \frac{(n+1)!n^{10}ln(n)}{n!(n+1)^{10}ln(n+1)} = \frac{n^{10}ln(n)}{(n+1)^{9}ln(n+1)}$

$\lim_{n\rightarrow \infty} \frac{n^{10}ln(n)}{(n+1)^{9}ln(n+1)}= \infty$

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Example Question #41 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{20n!}{n^{6}ln(n)}\newline note: \hspace f(n) = \frac{20n!}{n^{6}ln(n)}$

Possible Answers:

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{6}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 6$

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

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

Correct answer:

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

Explanation:

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

$\frac{\frac{\left(n+1\right)!}{\left(n+1\right)^6\ln \left(\left(n+1\right)\right)}}{\frac{n!}{n^6\ln \left(n\right)}} = \frac{(n+1)!n^6ln(n)}{n!(n+1)^6ln(n+1)} = \frac{(n+1)n^6ln(n)}{(n+1)^6ln(n+1)}$

$\lim_{n\rightarrow \infty} \frac{(n+1)n^6ln(n)}{(n+1)^6ln(n+1)}= \infty$

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Example Question #82 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{51*3^{n}n!}{2*n^{n}*e^{n}}\newline note: \hspace f(n) = \frac{51*3^{n}n!}{2*n^{n}*e^{n}}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e^2}{3}$

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{3}{e^2 }$

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

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{3}{e^2 }$

Explanation:

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

$\frac{\frac{3^{\left(n+1\right)}\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}e^{n+1}}}{\frac{3^nn!}{n^ne^n}} = \frac{3^{n+1}(n+1)!n^{n}e^{n}}{3^{n}n!(n+1)^{n+1}e^{n+1}} = \frac{3n^n}{(n+1)^{n}e}$

$\lim_{n\rightarrow \infty} \frac{3n^n}{(n+1)^{n}e} = \frac{3}{e^2}$

$\newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

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Example Question #41 : Convergence And Divergence

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{11*4^{n}n!}{16*n^{n}*e^{n}}\newline note: \hspace f(n) = \frac{11*4^{n}n!}{16*n^{n}*e^{n}}$

Possible Answers:

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e^2}{4}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{e^2}$

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

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{e^2}$

Explanation:

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

$\frac{\frac{4^{\left(n+1\right)}\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}e^{n+1}}}{\frac{4^nn!}{n^ne^n}} = \frac{4^{n+1}(n+1)!e^{n}n^{n})}{4^{n}n!e^{n+1}(n+1)^{n+1}} = \frac{4n^n}{e(n+1)^n}$

$\lim_{n\rightarrow \infty} \frac{4n^n}{e^n(n+1)^n} = \frac{4}{e^2}$

$\newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

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Example Question #41 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{25*5^{n}*n!}{n^{n}e^n}\newline note: \hspace f(n) = \frac{25*5^{n}*n!}{n^{n}e^n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{e^2}$

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

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e^2}{5}$

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

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

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{e^2}$

Explanation:

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

$\frac{\frac{5^{\left(n+1\right)}\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}e^{n+1}}}{\frac{5^nn!}{n^ne^n}} = \frac{5^{n+1}(n+1)!n^ne^n}{5^{n}n!(n+1)^{n+1}e^{n+1}} = \frac{5n^n}{(n+1)^ne}$

$\lim_{n\rightarrow \infty} \frac{5n^n}{(n+1)^ne} = \frac{5}{e^2}$

$\newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

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Example Question #86 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{36*6^{n}*n!}{n^{n}e^n}\newline note: \hspace f(n) = \frac{36*6^{n}*n!}{n^{n}e^n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{6}{e^2}$

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

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

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

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e^2}{6}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{6}{e^2}$

Explanation:

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

$\frac{\frac{6^{\left(n+1\right)}\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}e^{n+1}}}{\frac{6^nn!}{n^ne^n}} = \frac{6^{n+1}(n+1)!n^ne^n}{6^nn!(n+1)^{n+1}e^{n+1}} = \frac{6n^n}{(n+1)^ne}$

$\lim_{n\rightarrow \infty}\frac{6n^{n}}{(n+1)^{n}e} = \frac{6}{e^2}$

$\newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

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Example Question #41 : Convergence And Divergence

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{7^{n}*n!}{21n^{n}}\newline note: \hspace f(n) = \frac{7^{n}*n!}{21n^{n}}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$\frac{7^{\left(n+1\right)}\frac{\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}}}{7^n\frac{n!}{n^n}}= \frac{7n^{n}}{(n+1)^{n}}$

$\\\lim_{n\rightarrow \infty}\frac{7n^{n}}{(n+1)^{n}} = \frac{7}{e} \\ \newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

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Calculus 2 : Ratio Test

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2861 : Calculus Ii

Example Question #2861 : Calculus Ii

Example Question #2861 : Calculus Ii

Example Question #41 : Ratio Test

Example Question #41 : Ratio Test

Example Question #82 : Series In Calculus

Example Question #41 : Convergence And Divergence

Example Question #41 : Ratio Test

Example Question #86 : Series In Calculus

Example Question #41 : Convergence And Divergence

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