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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 #5 : Harmonic Series

Does the following series converge?

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

Possible Answers:

Cannot be determined

Yes

Correct answer:

Explanation:

No the series does not converge. The given problem is the harmonic series, which diverges to infinity.

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Example Question #6 : Harmonic Series

Does the following series converge?

$\sum_{n=1}^{\infty}\frac{cos(n\pi)}{n}$

Possible Answers:

Yes

Cannot be determined

Correct answer:

Yes

Explanation:

The series converges. The given problem is the alternating harmonic series, which converges by the alternating series test.

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Example Question #7 : Harmonic Series

Which of the following tests can be used to (successfully) test for the convergence/divergence of the harmonic series?

Possible Answers:

The Limit Test for Divergence

The Root Test

None of the given tests can be used.

The Ratio Test

The Integral Test

Correct answer:

The Integral Test

Explanation:

Only the Integral Test will work on the Harmonic Series, $\sum_{n=1}^\infty \frac{1}{n}$ .

To use the Integral Test, we evaluate

$\int_1^\infty \frac{1}{x}dx = [\ln(x)]_1^\infty = \lim_{x \to \infty}\ln(x)-0 =\infty$ , which shows that the series diverges.

Since $\frac{1}{n} \rightarrow 0$ , the Limit Test for Divergence fails.

The Ratio Test and the Root Test will always yield the same conclusion, so if one test fails, the both fail and vise versa.

For the Ratio Test,

$\lim_{x \to \infty}|\frac{1}{n+1}\times \frac{n}{1}| = \lim_{x \to \infty}\frac{n}{n+1} = 1$ . Since the result of the limit is , both tests fail.

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Example Question #8 : Harmonic Series

Let's say you are given harmonic series in the following form:

$\sum_{n=1}^{\infty}\frac{1}{n^p}$ ;

You are then asked to determine if the series converges, or diverges. For what values of p would this series be convergent? Assume p>0.

Possible Answers:

$p\in \mathbb{R}$

$p\leq1$

Correct answer:

Explanation:

The given series is called generalized harmonic series.

The series converges, if , and diverges, if $p\leq1$ .

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Example Question #1 : Alternating Series

By definition, an Alternating Series is a series of the form-

Possible Answers:

None of the other answers

$\sum_{k=k_0}^{\infty} (-1)^k \cdot a_k$

$\sum_{k=k_0}^{\infty} [(-1)^k\cdot a_{k+1} + (-1)^{k+1} \cdot a_k]$

$\sum_{k=k_0}^{\infty} (-1)^k + a_k \cdot a_{k+1}$

Correct answer:

$\sum_{k=k_0}^{\infty} (-1)^k \cdot a_k$

Explanation:

This type of series we can frequently check for convergence/divergence using the Alternating Series Test.

$\sum_{n=1}^\infty(-1)^{n+1}a_n=a_1-a_2+a_3-a_4+...$

The terms with an odd value for become negative since (-1)^n=-1 and the terms with an even value for are positive. This creates the alternating signs to occur within the sum.

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Example Question #3001 : Calculus Ii

Differentiate the following function.

$f(x)=x^4+\frac{x}{3x^2}+9$

Possible Answers:

$f'(x)=4x^3-\frac{1}{3x^2}$

f'(x)=x^3+999

$f'(x)=\frac{1}{4}x^2+3x$

$f'(x)=4x^3+\frac{1}{3x^3}$

Correct answer:

$f'(x)=4x^3-\frac{1}{3x^2}$

Explanation:

To differentiate the function we will need to use the Power Rule which states:

$f(x)=ax^n \rightarrow f'(x)=nax^{n-1}$

Looking at our function we can first simplify the equation.

$f(x)=x^4+\frac{x}{3x^2}+9=x^4+\frac{1}{3x}+9$

Applying the Power Rule we get:

$f'(x)=4x^3-\frac{1}{3x^2}$

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Example Question #1 : Alternating Series

Does the series $\sum_{n=0}^{\infty }\frac{(-1)^n}{n}$ converge conditionally, absolutely, or diverge?

Possible Answers:

Converge Conditionally.

Cannot tell with the given information.

Diverges.

Does not exist.

Converge Absolutely.

Correct answer:

Converge Conditionally.

Explanation:

The series converges conditionally.

The absolute values of the series $\sum_{n=0}^{\infty }\left |\frac{(1)^n}{n} \right |$ is a divergent p-series with $p\leq 1$ .

However, the the limit of the sequence $\lim_{n\rightarrow \infty }\frac{(-1)^n}{n}=0$ and it is a decreasing sequence.

Therefore, by the alternating series test, the series converges conditionally.

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Example Question #1 : Alternating Series

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Possible Answers:

$\left ( -\infty , \infty \right )$

$\left [ \frac{-1}{3}, \frac{1}{3} \right ]$

$\left ( \frac{-1}{3}, \frac{1}{3} \right )$

$\left ( -3, 3\right )$

(-1,0)

Correct answer:

$\left ( -\infty , \infty \right )$

Explanation:

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

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Example Question #2 : Alternating Series

Determine whether

$\small \sum_{n=1}^{\infty} (-1)^n \sin\frac{1}{n}$

converges or diverges, and explain why.

Possible Answers:

Divergent, by the test for divergence.

Convergent, by the alternating series test.

Divergent, by the comparison test.

More tests are needed.

Convergent, by the $\small p$ -series test.

Correct answer:

Convergent, by the alternating series test.

Explanation:

We can use the alternating series test to show that

$\small \sum_{n=1}^{\infty} (-1)^n \sin\frac{1}{n}$

converges.

We must have $\small \sin \frac{1}{n}\geq 0$ for $\small n\geq 1$ in order to use this test. This is easy to see because $\small \frac{1}{n}$ is in $\small \small \left[0,\frac{\pi}{2}\right]$ for all $\small n\geq 1$ (the values of this sequence are $\small 1,1/2,1/3,1/4,...$ ), and sine is always nonzero whenever sine's argument is in $\small \small \left[0,\frac{\pi}{2}\right]$ .

Now we must show that

1. $\small \small \lim_{n\to\infty} \sin \frac{1}{n}=0$

2. $\small \sin \frac{1}{n}$ is a decreasing sequence.

The limit

$\small \lim_{n\to\infty}\frac{1}{n}=0$

implies that

$\small \small \small \lim_{n\to\infty} \sin \frac{1}{n}=\sin 0=0$

so the first condition is satisfied.

We can show that $\small \small \small \sin \frac{1}{n}$ is decreasing by taking its derivative and showing that it is less than $\small 0$ for $\small n\geq 1$ :

$\small \small \small \small \small \frac{d}{dn}\sin \frac{1}{n}=-\frac{1}{n^2}\cos\frac{1}{n}$

The derivative is less than $\small 0$ , because $\small -\frac{1}{n^2}$ is always less than $\small 0$ , and that $\small \cos \frac{1}{n}$ is positive for $\small n\geq 1$ , using a similar argument we used to prove that $\small \small \sin \frac{1}{n}\geq 0$ for $\small n\geq 1$ . Since the derivative is less than $\small 0$ , $\small \small \small \sin \frac{1}{n}$ is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that

$\small \sum_{n=1}^{\infty} (-1)^n \sin\frac{1}{n}$

converges, by the alternating series test.

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Example Question #3 : Alternating Series

For the series: $\sum_{n=0}^{\infty} (-1)^n (\frac{n^n}{8^n})$ , determine if the series converge or diverge. If it diverges, choose the best reason.

Possible Answers:

$\textup{Diverges, by the Geometric Series Test.}$

$\textup{Diverge, since}\lim_{n \to \infty}x_n\neq 0.$

$\textup{Converge, since}\lim_{n \to \infty}x_n\neq 0.$

$\textup{Converge, since all Alternating Series Test rules are satisfied.}$

$\textup{Diverges, since the terms are not progressively decreasing}.$

Correct answer:

$\textup{Diverge, since}\lim_{n \to \infty}x_n\neq 0.$

Explanation:

The series given is an alternating series.

Write the three rules that are used to satisfy convergence in an alternating series test.

For $\sum_{n=1}^{\infty} (-1)^{n+1} x_n= x_1-x_2+x_3-x_4+...$ :

$\textup{1. All the } x_n \textup{ terms are positive.}$

$\textup{2. The terms must be decreasing: } x_n \geq x_{n+1}$

$\textup{3. The limit }\lim_{n \to \infty}x_n=0.$

$\sum_{n=0}^{\infty} (-1)^n (\frac{n^n}{8^n})=\sum_{n=0}^{\infty} (-1)^n (\frac{n}{8})^n$

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since $\lim_{n \to \infty}x_n\neq 0$ and instead approaches infinity.

The correct answer is:

$\textup{Diverge, since}\lim_{n \to \infty}x_n\neq 0.$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #5 : Harmonic Series

Example Question #6 : Harmonic Series

Example Question #7 : Harmonic Series

Example Question #8 : Harmonic Series

Example Question #1 : Alternating Series

Example Question #3001 : Calculus Ii

Example Question #1 : Alternating Series

Example Question #1 : Alternating Series

Example Question #2 : Alternating Series

Example Question #3 : Alternating Series

All Calculus 2 Resources

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