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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #41 : Series In Calculus

Let the series $\sum_{n=1}^{\infty}a_{n}$ with $a_{n}=\frac{(-1)^{n+1}}{n}$ determine whether the series is convergent or divergent using the ratio test.

Possible Answers:

We can't conclude when using the ratio test.

$\frac{-1}{ln2}$

We cant use the ratio test for this type of series.

The series is divergent.

The series is convergent.

Correct answer:

We can't conclude when using the ratio test.

Explanation:

Note that the series is alternating. To be able to use the ratio test, we will have to compute the ratio:

$\left|\frac{a_{n+1}}{a_n}\ \right|=\frac{(-1)^{n+2}}{n+1}\cdot \frac{n}{(-1)^{n+1}}=\frac{n}{n+1}$

Now we need to see that :

$\lim_{n\rightarrow \infty }\left|\frac{a_{n+1}}{a_n}\ \right|=\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$ .

Since $\lim_{n\rightarrow \infty }\left|\frac{a_{n+1}}{a_n}\ \right|=1$ we can't conclude by using the ratio test. We will have to call upon another test to show that the series is convergent.

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Example Question #12 : Ratio Test

Determine if the statement is true or false.

Assume that the series has positive terms. Furthermore suppose that

$\lim _{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=1$ , then all the series of the form $\sum_{n=1}^{\infty}a_n$ are divergent.

Possible Answers:

We can't conclude in general.

The series is divergent.

The series is convergent if it is positive.

There is only one series that satisfies that.

The statement above is false.

Correct answer:

The statement above is false.

Explanation:

To show that the statement above is false, we will consider the following example.

consider the series $\sum_{n=1}^{\infty}a_{n}$ where $a_{n}$ is given by:

$a_{n}=\frac{1}{n(n+1)}$ clearly the series has positive terms. Furthermore, we have

$\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=\frac{1}{(n+1)(n+2)}{\cdot \frac{n(n+1)}{1}}=\frac{n}{n+2}=1$ , meaning the series can be either convergent or divergent.

However, the series : $\sum_{n=1}^{\infty}a_{n}$ is convergent (use for example the integral test to see that it is convergent).

Therefore, the original statement is false.

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Example Question #11 : Ratio Test

We consider the series,

$\sum_{n=1}^{\infty}\frac{a^n}{n^b},0<a<\frac{1}{999} ,b>0$ .

Using the ratio test, what can we conclude about the nature of convergence of this series?

Possible Answers:

We will need to know the values of to decide.

The series is convergent.

We can't use the ratio test here.

The series is divergent.

The series converges to $\frac{a}{b}$ .

Correct answer:

The series is convergent.

Explanation:

Note that the series is positive.

As it is required we will use the ratio test to check for the nature of the series.

We have $\frac{a_{n+1}}{a_n}=\frac{a^{n+1}}{(n+1)^b}\frac{n^b}{a^n}=a\left(\frac{n}{n+1}\right)^b$ .

Therefore, $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty} a(\frac{n}{n+1})b$ $\lim_{n\rightarrow\infty}\frac{n^b}{(n+1)^b}=a(\lim_{n\rightarrow}\frac{n}{n+1})^b=a$

$L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L>1 the series diverges, if L<1 the series converges absolutely, and if L=1 the series may either converge or diverge.

Since $a<\frac{1}{999}<1$ the ratio test concludes that the series converges absolutely.

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Example Question #13 : Ratio Test

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:

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

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

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

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

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

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 #14 : Ratio Test

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:

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

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

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

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

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

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 #11 : Ratio Test

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:

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

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

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

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

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

Correct answer:

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

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 #16 : Ratio Test

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:

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

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

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

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

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

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 #17 : Ratio Test

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:

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

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

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

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

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

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 #18 : Ratio Test

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:

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

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

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

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

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

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 #19 : Ratio Test

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:

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

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

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

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

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

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

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

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Series In Calculus

Example Question #12 : Ratio Test

Example Question #11 : Ratio Test

Example Question #13 : Ratio Test

Example Question #14 : Ratio Test

Example Question #11 : Ratio Test

Example Question #16 : Ratio Test

Example Question #17 : Ratio Test

Example Question #18 : Ratio Test

Example Question #19 : Ratio Test

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