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 #4 : Ratio Test And Comparing Series

Determine if the following series is convergent, divergent or neither.

$\sum_{n=1}^{\infty} \frac{n!\ 4^n}{n+1}$

Possible Answers:

Divergent

Convergent

Neither

More tests are needed.

Inconclusive

Correct answer:

Divergent

Explanation:

To determine if

$\sum_{n=1}^{\infty} \frac{n!\ 4^n}{n+1}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

L<1 the series is absolutely convergent (and therefore convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{n!\ 4^n}{n+1}$

and

$a_{n+1}=\frac{(n+1)!\ 4^{n+1}}{n+2}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)!\ 4^{n+1}}{n+2}}{\frac{n!\ 4^n}{n+1}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)(n+1)!\ 4^{n+1}}{(n+2)n!\ 4^n}\right |$

Now lets simplify this.

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)(n+1) 4}{(n+2)}\right |$

When we evaluate the limit, we get.

$L=\infty$ .

Since L>1 , we have sufficient evidence to conclude that the series diverges.

Report an Error

Example Question #5 : Ratio Test And Comparing Series

Determine if the following series is divergent, convergent or neither.

$\sum_{n=0}^{\infty} \frac{n \ 2^n}{5^n}$

Possible Answers:

More tests are needed.

Divergent

Inconclusive

Neither

Convegent

Correct answer:

Convegent

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{n \ 2^n}{5^n}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

L<1 the series is absolutely convergent (and thus convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{n \ 2^n}{5^n}$

and

$a_{n+1}=\frac{(n+1) \ 2^{n+1}}{5^{n+1}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1) \ 2^{n+1}}{5^{n+1}}}{\frac{n \ 2^n}{5^n}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)\ 2^{n+1}\ 5^n}{n\ 2^n \ 5^{n+1}}\right |$ .

Now lets simplify this.

$L=\lim_{n\rightarrow \infty}\left | \frac{2(n+1)}{5n}\right |$

When we evaluate the limit, we get.

$L=\frac{2}{5}$ .

Since L<1 , we have sufficient evidence to conclude that the series converges.

Report an Error

Example Question #6 : Ratio Test And Comparing Series

Determine if the following series is convergent, divergent or neither.

$\sum_{n=1}^{\infty} \frac{n(-100)^n}{(n+1)5^{n+2}}$

Possible Answers:

Neither

Inconclusive

Convergent

Divergent

More tests needed.

Correct answer:

Divergent

Explanation:

To determine if

$\sum_{n=1}^{\infty} \frac{n(-100)^n}{(n+1)5^{n+2}}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

L<1 the series is absolutely convergent (therefore convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{n(-100)^n}{(n+1)5^{n+2}}$

and

$a_{n+1}=\frac{(n+1)(-100)^{n+1}}{(n+2)5^{n+3}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)(-100)^{n+1}}{(n+2)5^{n+3}}}{\frac{n(-100)^n}{(n+1)5^{n+2}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2(-100)^{n+1}{5^{n+2}}}{n(n+2)\ (-100)^n \ 5^{n+3}}\right |$

Now lets simplify this.

$L=20\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |$

When we evaluate the limit, we get.

L=20 .

Since L>1 , we have sufficient evidence to conclude that the series diverges.

Report an Error

Example Question #7 : Ratio Test And Comparing Series

Determine if the following series is divergent, convergent or neither.

$\sum_{n=0}^{\infty} 4^{2n+1}n!$

Possible Answers:

Inconclusive

Neither

More tests are needed.

Divergent

Convergent

Correct answer:

Divergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} 4^{2n+1}n!$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

L<1 the series is absolutely convergent (and thus convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n= 4^{2n+1}n!$

and

$a_{n+1}= 4^{2(n+1)+1}(n+1)!=4^{2n+3}(n+1)!$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{4^{2n+3}(n+1)!}{4^{2n+1}n!}\right |$

We can simplify the expression to be

$L=\lim_{n\rightarrow \infty}\left | 16 (n+1)}\right |$

When we evaluate the limit, we get.

$L=\infty$ .

Since L>1 , we have sufficient evidence to conclude that the series diverges.

Report an Error

Example Question #8 : Ratio Test And Comparing Series

Determine of the following series is convergent, divergent or neither.

$\sum_{n=0}^{\infty} \frac{n4^{2n}}{(n+1)3^{2n}}$

Possible Answers:

Divergent

Inconclusive.

Convergent

Neither

More tests are needed.

Correct answer:

Divergent

Explanation:

To determine whether this series is convergent, divergent or neither

$\sum_{n=0}^{\infty} \frac{n4^{2n}}{(n+1)3^{2n}}$

we need to remember the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

L<1 the series is absolutely convergent (and therefore convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n= \frac{n4^{2n}}{(n+1)3^{2n}}$

and

$a_{n+1}= \frac{(n+1)4^{2(n+1)}}{(n+2)3^{2(n+1)}}= \frac{(n+1)4^{2n+2}}{(n+2)3^{2n+2}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)4^{2n+2}}{(n+2)3^{2n+2}}}{\frac{n4^{2n}}{(n+1)3^{2n}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2\ 3^{2n} \ 4^{2n+2}}{n(n+2)\ 3^{2n+2}\ 4^{2n}}\right |$

Now lets simplify this to.

$L=\frac{16}{9}\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |$

When we evaluate the limit, we get.

$L=\frac{16}{9}$ .

Since L>1 , we have sufficient evidence to conclude that the series is divergent.

Report an Error

Example Question #9 : Ratio Test And Comparing Series

Determine what the following series converges to using the ratio test and whether the series is convergent, divergent or neither.

$\sum_{n=0}^{\infty} \frac{n2^{4n}}{(n+1)5^{6n}}$

Possible Answers:

$L=\frac{16}{15625}$ , and convergent.

$L=\frac{16}{15625}$ , and divergent.

$L=\frac{16}{15625}$ , and neither.

L=1 , and neither.

L=1 , and convergent.

Correct answer:

$L=\frac{16}{15625}$ , and convergent.

Explanation:

To determine whether this series is convergent, divergent or neither

$\sum_{n=0}^{\infty} \frac{n2^{4n}}{(n+1)5^{6n}}$

we need to remember the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

L<1 the series is absolutely convergent (thus convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n= \frac{n2^{4n}}{(n+1)5^{6n}}$

and

$a_{n+1}= \frac{(n+1)2^{4(n+1)}}{(n+2)5^{6(n+1)}}= \frac{(n+1)2^{4n+4}}{(n+2)5^{6n+6}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)2^{4n+4}}{(n+2)5^{6n+6}}}{\frac{n2^{4n}}{(n+1)5^{6n}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2\ 2^{4n+4} \ 5^{6n}}{n(n+2)\ 2^{4n}\ 5^{6n+6}}\right |$

Now lets simplify this to.

$L=\frac{16}{15625}\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |$

When we evaluate the limit, we get.

$L=\frac{16}{15625}$ .

Since L<1 , we have sufficient evidence to conclude that the series is convergent.

Report an Error

Example Question #10 : Ratio Test And Comparing Series

Determine the convergence or divergence of the following series:

$\sum_{n=0}^{\infty }\frac{2^n}{n!}$

Possible Answers:

The series is divergent.

The series is conditionally convergent.

The series may be divergent, conditionally convergent, or absolutely convergent.

The series (absolutely) convergent.

Correct answer:

The series (absolutely) convergent.

Explanation:

To determine the convergence or divergence of this series, we use the Ratio Test:

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

If L<1 , then the series is absolutely convergent (convergent)

If L>1 , then the series is divergent

If L=1 , the series may be divergent, conditionally convergent, or absolutely convergent

So, we evaluate the limit according to the formula above:

$\lim_{n\rightarrow \infty }\left | \frac{2^{n+1}}{(n+1)!}\cdot \frac{n!}{2^n}\right |$

which simplified becomes

$\lim_{n\rightarrow \infty }\left | \frac{2^{n+1}}{(n+1)(n!)}\cdot \frac{(n!)}{2^n}\right |$

Further simplification results in

$\lim_{n\rightarrow \infty }\left |\frac{2}{n+1} \right |=0<1$

Therefore, the series is absolutely convergent.

Report an Error

Example Question #71 : Ratio Test

Determine the convergence or divergence of the following series:

$\sum_{n=0}^{\infty }\frac{3^n(n+4)}{n!}$

Possible Answers:

The series is (absolutely) convergent.

The series may be divergent, conditionally convergent, or absolutely convergent.

The series is divergent.

The series is conditionally convergent.

Correct answer:

The series is (absolutely) convergent.

Explanation:

To determine the convergence or divergence of this series, we use the Ratio Test:

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

If L<1 , then the series is absolutely convergent (convergent)

If L>1 , then the series is divergent

If L=1 , the series may be divergent, conditionally convergent, or absolutely convergent

So, follow the above formula:

$\lim_{n\rightarrow \infty }\left | \frac{3^{n+1}(n+5)}{(n+1)!}\cdot \frac{(n!)}{3^n(n+4)}\right |$

Now simplify and evaluate the limit:

$\lim_{n\rightarrow \infty }\frac{3(n+5)}{n(n+4)}=\lim_{n\rightarrow \infty }(\frac{3n+15}{n^2+4n})=0<1$

Because the limit is less than one, the series is absolutely convergent.

Report an Error

Example Question #111 : Series In Calculus

Using the Ratio Test, determine what the following series converges to, and whether the series is Divergent, Convergent or Neither.

$\sum_{n=0}^{\infty} \frac{2n\ 4^n}{(n+1)\ 2^n}$

Possible Answers:

$L=\infty$ , and Divergent

L=2 , and Convergent

$L=\infty$ , and Neither

L=2 , and Neither

L=2 , and Divergent

Correct answer:

L=2 , and Divergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{2n\ 4^n}{(n+1)\ 2^n}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

L<1 the series is absolutely convergent (and hence convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n= \frac{2n\ 4^n}{(n+1)\ 2^n}$

and

$a_{n+1}= \frac{2(n+1)\ 4^{n+1}}{(n+2)\ 2^{n+1}}=\frac{(2n+2)\ 4^{n+1}}{(n+2) \ 2^{n+1}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(2n+2)\ 4^{n+1}}{(n+2) \ 2^{n+1}}}{ \frac{2n\ 4^n}{(n+1)\ 2^n}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(2n+2)(n+1)\ 4^{n+1}\ 2^n}{2n(n+2)4^n\ 2^{n+1}}\right |$

We can simplify the expression to

$L=2\lim_{n\rightarrow \infty}\left |\frac{(2n+2)(n+1)}{2n(n+2)} \right |$

When we evaluate the limit, we get.

L=2 .

Since L>1 , we have sufficient evidence to conclude that the series is Divergent.

Report an Error

Example Question #112 : Series In Calculus

Determine what the following series converges to, and whether the series is Convergent, Divergent or Neither.

$\sum_{n=0}^{\infty} \frac{n^2\ 2^n}{(n+1)\ 3^n}$

Possible Answers:

$L=\frac{2}{3}$ , and Divergent

$L=\frac{2}{3}$ , and Neither

$L=\frac{2}{3}$ , and Convergent

L=1 , and Neither

$L=\infty$ , and Convergent

Correct answer:

$L=\frac{2}{3}$ , and Convergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{n^2\ 2^n}{(n+1)\ 3^n}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

L<1 the series is absolutely convergent (and hence convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

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

and

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

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)^2\ 2^{n+1}}{(n+2)\ 3^{n+1}}}{ \frac{n^2\ 2^n}{(n+1)\ 3^n}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^3\ 2^{n+1}\ 3^n}{n^2(n+2)\ 2^n\ 3^{n+1}}\right |$

We can simplify the expression to

$L=\frac{2}{3}\lim_{n\rightarrow \infty}\left |\frac{(n+1)^3}{n^2(n+2)} \right |$

When we evaluate the limit, we get.

$L=\frac{2}{3}$ .

Since L<1 , we have sufficient evidence to conclude that the series is Convergent.

Report an Error

View Calculus 2 Tutors

Dalton
Certified Tutor

University of Massachusetts Amherst, Doctor of Philosophy, Mathematics.

View Calculus 2 Tutors

Vandana
Certified Tutor

Jawaharlal Nehru Technological University, Bachelor of Technology, Civil Engineering. Florida International University, Maste...

View Calculus 2 Tutors

Bobby
Certified Tutor

University of Houston - Downtown, Bachelor of Science, Applied Mathematics.

All Calculus 2 Resources

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

Popular Subjects

Statistics Tutors in Atlanta, Reading Tutors in Chicago, French Tutors in Philadelphia, English Tutors in Boston, ACT Tutors in Dallas Fort Worth, French Tutors in Phoenix, MCAT Tutors in Dallas Fort Worth, GMAT Tutors in Atlanta, French Tutors in San Diego, SSAT Tutors in Houston

Popular Courses & Classes

SSAT Courses & Classes in Houston, Spanish Courses & Classes in Chicago, SSAT Courses & Classes in San Diego, ISEE Courses & Classes in Washington DC, LSAT Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Atlanta, ACT Courses & Classes in Boston, GMAT Courses & Classes in Denver, ISEE Courses & Classes in Houston

Popular Test Prep

LSAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Houston, ISEE Test Prep in Philadelphia, LSAT Test Prep in Boston, GRE Test Prep in Phoenix, MCAT Test Prep in San Diego, ACT Test Prep in Boston, SSAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Miami, SSAT Test Prep in Seattle

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 #4 : Ratio Test And Comparing Series

Example Question #5 : Ratio Test And Comparing Series

Example Question #6 : Ratio Test And Comparing Series

Example Question #7 : Ratio Test And Comparing Series

Example Question #8 : Ratio Test And Comparing Series

Example Question #9 : Ratio Test And Comparing Series

Example Question #10 : Ratio Test And Comparing Series

Example Question #71 : Ratio Test

Example Question #111 : Series In Calculus

Example Question #112 : Series In Calculus

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: