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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #84 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{8^{n}*n!}{32n^{n}}\newline note: \hspace f(n) = \frac{8^{n}*n!}{32n^{n}}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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Example Question #89 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{9^{n}*n!}{81n^{n}}\newline note: \hspace f(n) = \frac{9^{n}*n!}{81n^{n}}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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Example Question #90 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{10^{n}*n!}{99n^{n}}\newline note: \hspace f(n) = \frac{10^{n}*n!}{99n^{n}}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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Example Question #91 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{4n!}{3^n\left(n-1\right)!} \newline note: \hspace f(n) = \frac{4n!}{3^n\left(n-1\right)!}$

Possible Answers:

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

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

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

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

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{3}$

Correct answer:

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

Explanation:

Determine the convergence of the series based on the limits.

$\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

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

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

$\lim_{n\rightarrow \infty} \frac{\left(n+1\right)!\left(n-1\right)!}{3n!^2} = \frac{1}{3}$

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Example Question #92 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{3^n\cdot \:2n!}{3\left(n-1\right)!} \newline note: \hspace f(n) = \frac{3^n\cdot \:2n!}{3\left(n-1\right)!}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

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

$= \frac{3^{n+1}\left(n+1\right)!\left(n-1\right)!}{3^nn!^2} = \frac{3\left(n+1\right)!\left(n-1\right)!}{n!^2}$

$\lim_{n\rightarrow \infty} \frac{3\left(n+1\right)!\left(n-1\right)!}{n!^2} = 3$

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Example Question #93 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{2^n}{(n-10)!} \newline note: \hspace f(n) = \frac{2^n}{(n-10)!}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{2^{(n+1)}}{((n+1)-10)!}} {\frac{2^n}{(n-10)!}}$ $= \frac{\frac{2^{(n+1)}}{2^{n}}}{\frac{(n+1-10)!}{(n-10)!}}$ $= \frac{2}^{n-9}$ $, \lim_{n\rightarrow \infty} \frac{2}{n-9} = 0$

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Example Question #94 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{2^n}{(n-7)!} \newline note: \hspace f(n) = \frac{2^n}{(n-7)!}$

Possible Answers:

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{2}{7}$

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

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

Correct answer:

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

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{2^{n+1}}{((n+1)-7)!}} {\frac{2^{n}}{(n-7)!}}$ $= \frac{\frac{2^{(n+1)}}{2^{n}}}{\frac{(n+1-7)!}{(n-7)!}}$ $= \frac{2}{n-6}$ $, \lim_{n\rightarrow \infty} \frac{2}{n-6} = 0$

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Example Question #95 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{4^{n}}{(2n-1)!} \newline note: \hspace f(n) = \frac{4^{n}}{(2n-1)!}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{4^{(n+1)}}{(2(n+1)-1)!}} {\frac{4^{n}}{(2n-1)!}}$ $= \frac{\frac{4^{(n+1)}}{4^{n}}}{\frac{(n+1-1)!}{(n-1)!}}$ $= \frac{4}{n(2n+1)}$ $,\lim_{n\rightarrow \infty} \frac{4}{n(2n+1)} = 0$

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Example Question #51 : Convergence And Divergence

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{4^n}{(2n-7)!} \newline note: \hspace f(n) = \frac{4^n}{(2n-7)!}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{7}$

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

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

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

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

Correct answer:

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

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{4^{(n+1)}}{(2(n+1)-7)!}} {\frac{4^{n}}{(2n-7)!}}$ $= \frac{\frac{4^{(n+1)}}{4^{n}}}{\frac{(2n+2-7)!}{(2n-7)!}}$ $= \frac{2}{\left(n-3\right)\left(2n-5\right)}$ , $\lim _{n\to \infty \:}\left\frac{2}{\left(n-3\right)\left(2n-5\right)}\right\right = 0$

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Example Question #52 : Convergence And Divergence

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{50!}{5^n\left(n-1\right)!}\newline note: \hspace f(n) = \frac{50!}{5^n\left(n-1\right)!}$

Possible Answers:

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

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

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

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 50$

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

Correct answer:

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

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{1}{5^{n+1}\left(\left(n+1\right)-1\right)!}}{\frac{1}{5^n\left(n-1\right)!}}$ $= \frac{5^n\left(n-1\right)!}{5^{n+1}n!}$ $= \frac{1}{5n}$ $,\lim_{n\rightarrow \infty} \frac{1}{5n} = 0$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #84 : Series In Calculus

Example Question #89 : Series In Calculus

Example Question #90 : Series In Calculus

Example Question #91 : Series In Calculus

Example Question #92 : Series In Calculus

Example Question #93 : Series In Calculus

Example Question #94 : Series In Calculus

Example Question #95 : Series In Calculus

Example Question #51 : Convergence And Divergence

Example Question #52 : Convergence And Divergence

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