We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Convergence and Divergence

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

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

Example Questions

Example Question #1 : Ratio Test

Which of these series cannot be tested for convergence/divergence properly using the ratio test? (Which of these series fails the ratio test?)

Possible Answers:

None of the other answers.

$\sum_{k=0}^{\infty} \frac{(-2)^k}{k!}$

$\sum_{k=0}^{\infty} \frac{1}{k!}$

$\sum_{k=0}^{\infty} k$

$\sum_{k=0}^{\infty} \frac{k^2}{3^k}$

Correct answer:

$\sum_{k=0}^{\infty} k$

Explanation:

The ratio test fails when $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | =1$ . Otherwise the series converges absolutely if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | <1$ , and diverges if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | >1$ .

Testing the series $\sum_{k=0}^{\infty} k$ , we have

$\lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left |\frac{(k+1)}{(k)} \right | = \lim_{k\to\infty} 1+ \frac{1}{k} = 1.$

Hence the ratio test fails here. (It is likely obvious to the reader that this series diverges already. However, we must remember that all intuition in mathematics requires rigorous justification. We are attempting that here.)

Report an Error

Example Question #2 : Convergence And Divergence

Assuming that $a_{n}=n^{\alpha}(n+1)$ , $\alpha \ge 2$ . Using the ratio test, what can we say about the series:

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

Possible Answers:

$\infty$

We cannot conclude when we use the ratio test.

$-\infty$

It is convergent.

Correct answer:

We cannot conclude when we use the ratio test.

Explanation:

As required by this question we will have to use the ratio test. $L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

To do so, we will need to compute : $\frac{a_{n+1}}{a_{n}}$ . In our case:

$a_{n}=\frac{1}{a_{n}}$

Therefore

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}$ .

We know that $\lim_{n\rightarrow \infty }\frac{n}{n+1}=1$

This means that

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}=1\forall\quad \alpha$

Since L=1 by the ratio test, we can't conclude about the convergence of the series.

Report an Error

Example Question #1 : Ratio Test

We consider the series : $\sum_{n=1}^{\infty}\frac{1}{72n+89}$ , use the ratio test to determine the type of convergence of the series.

Possible Answers:

The series is fast convergent.

$\frac{1}{79}$

It is clearly divergent.

$\frac{3}{79}$

We cannot conclude about the nature of the series.

Correct answer:

We cannot conclude about the nature of the series.

Explanation:

To be able to use to conclude using the ratio test, we will need to first compute the ratio then use $L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge. Computing the ratio we get,

$\frac{a_{n+1}}{a_{n}}$ .

We have then:

$a_{n+1}=\frac{1}{72(n+1)+89}=\frac{1}{72n+161}$

Therefore have :

$\frac{a_{n+1}}{a_{n}}=$ $\frac{72n+89}{72n+161}$

It is clear that $\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=1$ .

By the ratio test , we can't conclude about the nature of the series.

Report an Error

Example Question #3 : Convergence And Divergence

Using the ratio test,

what can we say about the series.

$\sum_{_n=1}^{\infty} \frac{1}{n^p}$ where is an integer that satisfies:

p> 1

Possible Answers:

$\infty$

We can't conclude when we use the ratio test.

$\frac{4}{5}$

We can't use the ratio test to study this series.

Correct answer:

We can't conclude when we use the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

The Ratio Test states:

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

then if,

1) L<1 the series converges absolutely.

2) L>1 the series diverges.

3) L=1 the series either converges or diverges.

Therefore we need to evaluate,

$\frac{a_{n+1}}{a_{n}}$

we have,

$a_{n+1}=\frac{1}{(n+1)^p} ,a_{n}=\frac{1}{n^p}$

therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^p}{(n+1)^p}$ .

We know that

$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$

and therefore,

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

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

Report an Error

Example Question #2 : Ratio Test

Consider the following series :

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

$a_{n}=\frac{n}{n^\alpha +1},\alpha \ge 3$ . Using the ratio test, find the nature of the series.

Possible Answers:

The series is convergent.

$\frac{3}{211}$

$\frac{\pi}{e}$

We can't conclude when using the ratio test.

Correct answer:

We can't conclude when using the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

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

We need to evaluate,

$\frac{a_{n+1}}{a_{n}}$ we have:

$a_{n+1}= \frac{n+1}{(n+1)^\alpha +1},a_{n}=\frac{n}{n^\alpha +1}$ .

Therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^\alpha +1}{(n+1)^\alpha +1}\frac{n+1}{n}$ . We know that,

$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$ and therefore

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

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$ .

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

Report an Error

Example Question #1 : Ratio Test

Use the ratio test to determine whether the series below is convergent or divergent.

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

Possible Answers:

The series is convegent.

The series is divergent.

$\frac{4}{13}$

$\frac{6}{11}$

Correct answer:

The series is divergent.

Explanation:

To use the ratio test, we will need to compute the ratio

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

We have $a_{n+1}=\frac{(n+1)!}{2^{n+1}}, a_n=\frac{n!}{2^n}$ we have then :

$\frac{a_{n+1}}{a_n}=\frac{2^n(n+1)!}{2^{n+1}n!}$ .

Since we can write :

(n+1)!=(n+1)n!

$\frac{a_{n+1}}{a_n}=\frac{n+1}{2}$

Thus $\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=\lim_{n\rightarrow \infty} \frac{2^n(n+1)!}{2^{n+1}n!}=\infty$ and because $\infty >1$ the series must diverge.

Therefore we conclude that the series is divergent.

Report an Error

Example Question #1 : Ratio Test

Using the ratio test , what can you say about the following series:

$\sum_{n=1}^{\infty}\frac{sin(n)}{7^n}$

Possible Answers:

The series has two limits.

The series will converge and diverge when it gets close to $\infty$ .

The series is divergent.

The series is convergent.

$\frac{sin(1)}{3}$

Correct answer:

The series is convergent.

Explanation:

We will use the comparison test to conclude about the convergence of this series. To show that the majorant series is convergent we will have to call upon the ratio test.

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

We note first that,

$|sin(n)|\le 1 \forall n$ where n is positive integer.

We have $\left|\frac{sin(n)}{7^n}\right|\le\frac{1}{7^n}$ . By the comparison test, if we can show that the series $\sum_{n=1}^{\infty}\frac{1}{7^n}$ is convergent, then by the comparison test, the series $\sum_{n=1}^{\infty}\frac{sin(n)}{7^n}$ is also convergent.

We consider now the series : $\sum_{n=1}^{\infty}\frac{1}{7^n}$ . We have:

$\frac{a_{n+1}}{a_n}=\frac{1}{7}$ and since $\frac{1}{7}< 1$ we conclude that the series is convergent by the ratio test.

This shows that our series is convergent.

Report an Error

Example Question #2 : Convergence And Divergence

Suppose that a series has positive terms.

If $\frac{a_n}{a_{n+1}}=\frac{2^n}{3^{n+1}}$ what can we say about the convergency of the series.

Possible Answers:

The series is convergent.

We can't conclude.

The series is divergent.

We will need to know the explicit formula for $a_{n}$ .

We will need to know the first two terms.

Correct answer:

The series is divergent.

Explanation:

We are given that the series has positive terms. We know that

$\frac{a_n}{a_{n+1}}=\frac{2^n}{3^{n+1}}$ . This means that $\frac{a_{n+1}}{a_{n}}=3\frac{3^n}{2^n}}$ .

Now we note that $\lim_{n\rightarrow \infty }\frac{3^n}{2^n}=\infty$ .

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

Since $\infty>1$ the series will diverge.

The ratio test lets us conclude that the series is divergent.

Report an Error

Example Question #8 : Ratio Test

We will consider the following series :

$\sum_{n=2}^{\infty}\frac{1}{ln(n)^\alpha}$ .

What can you say about the nature of this series using the ratio test? Assume that $\alpha > 0$ .

Possible Answers:

The series is convergent.

We need to know the exact value of $\alpha$ .

The series converges to $\frac{1}{\alpha}$ .

The nature of the series depends on $\alpha$ .

We can't conclude about the nature of the series.

Correct answer:

We can't conclude about the nature of the series.

Explanation:

Note that for $n\ge 2$ and $\alpha>0$ the series is always positive.

To be able to use the ratio test, we will have to compute the ratio:

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

We have $a_{n+1}=\frac{1}{ln(n+1)^\alpha}$ hence :

$\frac{a_{n+1}}{a_{n}}=\frac{ln(n)^\alpha}{ln(n+1)^\alpha}=\left(\frac{ln(n)}{ln(n+1)}\right)^\alpha$

Therefore :

$\lim_{n\rightarrow \infty } \frac{a_{n+1}}{a_{n}}=1$ since $\lim_{n\rightarrow \infty }(\frac{ln(n)}{ln(n+1)})^\alpha=1$

By the ratio test we cannot conclude about the nature of the series.

Report an Error

Example Question #1 : Convergence And Divergence

We consider the following series:

$\sum_ {n=1}^{\infty}a_{n}$ where $a_{n}=\frac{1}{n\sqrt[n]{3}}$ .

Using the ratio test what can you say about the nature of the series?

Is it convergent or divergent?

Possible Answers:

The series is convergent.

We can't conclude using the ratio test.

$\frac{1}{2}$

$\frac{1}{3}$

The series is divergent.

Correct answer:

We can't conclude using the ratio test.

Explanation:

We will use the ratio test noting first that the series is positive.

We will compute the ratio:

$\frac{a_{n+1}}{a_n}$ . Note that: $a_{n+1}=\frac{1}{(n+1)\sqrt[n+1]{3}}$

Hence : $\frac{a_{n+1}}{a_n}=\frac{n}{n+1}\frac{\sqrt[n]{3}}{\sqrt[n+1]{3}}$

Now we have $\lim_{n\rightarrow \infty} \frac{n}{n+1}=1$ and $\frac{\sqrt[n]{3}}{\sqrt[n+1]{3}}=3^{{\frac{1}{n}}-\frac{1}{n+1}}=3^{\frac{1}{n(n+1)}}$

and we have $\lim_{n\rightarrow \infty}3^{\frac{1}{n(n+1)}}=\lim_{n\rightarrow \infty}e^{\frac{1}{n(n+1)}ln(3))}=1$

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

Therefore the ratio test is inconclusive. We will need to use another test .

Report an Error

View Calculus 2 Tutors

Timothy
Certified Tutor

Bethune-Cookman University, Bachelor of Science, Mathematics. University of Phoenix, Master of Science, Computer and Informat...

View Calculus 2 Tutors

Evgenii
Certified Tutor

View Calculus 2 Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

All Calculus 2 Resources

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

Popular Subjects

ISEE Tutors in Chicago, Reading Tutors in San Francisco-Bay Area, English Tutors in San Diego, MCAT Tutors in Washington DC, GMAT Tutors in Houston, Math Tutors in San Francisco-Bay Area, GRE Tutors in Chicago, SAT Tutors in Denver, Chemistry Tutors in San Francisco-Bay Area, Chemistry Tutors in Denver

Popular Courses & Classes

ACT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Houston, SAT Courses & Classes in San Diego, LSAT Courses & Classes in Denver, LSAT Courses & Classes in Atlanta, ACT Courses & Classes in Denver, ISEE Courses & Classes in Chicago, Spanish Courses & Classes in Washington DC, MCAT Courses & Classes in Chicago, LSAT Courses & Classes in Philadelphia

Popular Test Prep

ACT Test Prep in Houston, GRE Test Prep in Philadelphia, GRE Test Prep in Miami, MCAT Test Prep in Phoenix, LSAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Philadelphia, GRE Test Prep in New York City, GMAT Test Prep in Los Angeles, SSAT Test Prep in Phoenix, ACT Test Prep in Dallas Fort Worth

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : Convergence and Divergence

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Ratio Test

Example Question #2 : Convergence And Divergence

Example Question #1 : Ratio Test

Example Question #3 : Convergence And Divergence

Example Question #2 : Ratio Test

Example Question #1 : Ratio Test

Example Question #1 : Ratio Test

Example Question #2 : Convergence And Divergence

Example Question #8 : Ratio Test

Example Question #1 : Convergence And Divergence

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: