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Calculus 2 : Series in Calculus

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Example Questions

Example Question #51 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{6}{5^n}$

Note: $f(n) = \frac{6}{5^n}$

Possible Answers:

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.2$

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

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

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.2$

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{1}{5^{n+1}}}{\frac{1}{5^n}}$ $= \frac{5^n}{5^{n+1}}=0.2$

$\lim_{n\rightarrow \infty} 0.2 \rightarrow 0.2$

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Example Question #52 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{10}{2^n}$

Note: $f(n) = \frac{10}{2^n}$

Possible Answers:

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

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$

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

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$

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 $\lim_{n\rightarrow \infty}$

$\frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}}$ $= \frac{2^n}{2^{n+1}}=0.5$

$= \lim_{n\rightarrow \infty} 0.5 \rightarrow 0.5$

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Example Question #53 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{e^n}{3^n}$

Note: $f(n) = \frac{e^n}{3^n}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{e^{n+1}}{3^{n+1}}}{\frac{e^n}{3^n}}$ $= \frac{\frac{e^{n+1}}{e^n}}{\frac{3^{n+1}}{3^(n)}}$ $= \frac{e}^{3}$

$\lim_{n\rightarrow \infty} \frac{e}{3} \rightarrow \frac{e}{3}$

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Example Question #54 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{3^n}{2^n}$

Note: $f(n) = \frac{3^n}{2^n}$

Possible Answers:

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

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1.5$

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1.5$

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{3^{n+1}}{2^{n+1}}}{\frac{3^n}{2^n}}$ $= \frac{\frac{3^{n+1}}{3^n}}{\frac{2^{n+1}}{2^n}}=\frac{3}{2}$

$\frac{3}{2}= 1.5$

$=\lim_{n\rightarrow \infty} 1.5 \rightarrow 1.5$

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Example Question #55 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{e^n}{5^n}$

Note: $f(n) = \frac{e^n}{5^n}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{e^{n+1}}{5^{n+1}}}{\frac{e^n}{5^n}}$ $= \frac{\frac{e^{n+1}}{e^n}}{\frac{5^{n+1}}{5^n}}$ $= \frac{e}^{5}$

$\lim_{n\rightarrow \infty} \frac{e}{5} \rightarrow \frac{e}{5}$

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Example Question #56 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{10e^n}{4^n}$

Note: $f(n) = \frac{e^n}{4^n}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 10\frac{e}{4}$

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

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

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

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

Correct answer:

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

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{e^{n+1}}{4^{n+1}}}{\frac{e^n}{4^n}}$ $= \frac{\frac{e^{n+1}}{e^n}}{\frac{4^{n+1}}{4^n}}=\frac{e}{4}$ $,\lim_{n\rightarrow \infty} \frac{e}{4} \rightarrow$ $\frac{e}^{4}$

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Example Question #57 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{n!}{e^n}$

Note: $f(n) = \frac{n!}{e^n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.25$

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

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

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

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

Correct answer:

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

Explanation:

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 $\lim_{n\rightarrow \infty}$ .

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

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Example Question #58 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{(n-1)!}{e^n}$

Note: $f(n) = \frac{(n-1)!}{e^n}$

Possible Answers:

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

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.4$

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

Correct answer:

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

Explanation:

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 $\lim_{n\rightarrow \infty}$ .

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

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Example Question #59 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{7n!}{9n}$

Note: $f(n) = \frac{7n!}{9n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{9}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{9}{7}$

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

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

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

Correct answer:

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

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 $\lim_{n\rightarrow \infty}$

$\frac{\frac{(n+1)!}{(n+1)}} {\frac{n!}{n}}$ $= \frac{\frac{(n+1)!}{(n)!}}{\frac{n+1}{n}}$ $= n ,\lim_{n\rightarrow \infty} n \rightarrow \infty$

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Example Question #60 : Series In Calculus

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

$\sum_{n = 1}^{\infty} \frac{n!}{2n}$

Note: $f(n) = \frac{n!}{2n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$

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

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1.5$

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

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

Correct answer:

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

Explanation:

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 simplified, take $\lim_{n\rightarrow \infty}$

$\frac{\frac{(n+1)!}^{(n+1)}}^{\frac{n!}^{n}}$ $= \frac{(n+1)! \cdot n}{(n+1)\cdot n!}=n$

$=\lim_{n\rightarrow \infty} n \rightarrow \infty$

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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : Series In Calculus

Example Question #52 : Series In Calculus

Example Question #53 : Series In Calculus

Example Question #54 : Series In Calculus

Example Question #55 : Series In Calculus

Example Question #56 : Series In Calculus

Example Question #57 : Series In Calculus

Example Question #58 : Series In Calculus

Example Question #59 : Series In Calculus

Example Question #60 : Series In Calculus

All Calculus 2 Resources

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