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Calculus 2 : Series in Calculus

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Example Questions

Example Question #71 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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Example Question #72 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #73 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #74 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #75 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #76 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #77 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #78 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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Example Question #79 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #80 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Calculus 2 : Series in Calculus

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