We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Series in Calculus

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 #83 : Convergence And Divergence

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

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

Possible Answers:

Neither

Divergent

More tests are needed.

Not enough information.

Convergent

Correct answer:

Convergent

Explanation:

To determine if

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

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

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

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

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

Example Question #2902 : Calculus Ii

Determine if the following series is Convergent, Divergent or Neither.

$\sum_{n=0}^{\infty} \frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}$

Possible Answers:

Not enough information.

Neither

Convergent

More tests are needed.

Divergent

Correct answer:

Convergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}$

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 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

$a_n=\frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}$

and

$a_{n+1}=\frac{(2(n+1)+3)4^{n+3}}{(n+3)5^{n+2}}=\frac{(2n+5)4^{n+3}}{(n+3)5^{n+2}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(2n+5)4^{n+3}}{(n+3)5^{n+2}}}{\frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(2n+5)(n+2)4^{n+3}\ 5^{n+1}}{(n+3)(2n+3)5^{n+2}\ 4^{n+2}}\right |$

We can simplify the expression to

$L=\frac{4}{5}\lim_{n\rightarrow \infty}\left | \frac{(2n+5)(n+2)}{(n+3)(2n+3)}\right |$

When we evaluate the limit, we get.

$L=\frac{4}{5}$ .

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

Report an Error

Example Question #85 : Convergence And Divergence

Determine whether the following series is Convergent, Divergent or Neither.

$\sum_{n=0}^{\infty} \frac{n2^n}{n!}$

Possible Answers:

Divergent

Neither

Not enough information.

More tests are needed.

Convergent

Correct answer:

Convergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{n2^n}{n!}$

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

$a_n=\frac{n2^n}{n!}$

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

When we evaluate the limit, we get.

L=0 .

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

Report an Error

Example Question #121 : Series In Calculus

Determine whether the following series converges or diverges:

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

Possible Answers:

The series diverges.

The series (absolutely) converges.

The series conditionally converges.

The series may converge, diverge, or be conditionally convergent.

Correct answer:

The series (absolutely) converges.

Explanation:

To determine whether the given series is convergent or divergent, we must use the Ratio Test, which states that for

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

if L is less than 1, then the series is absolutely convergent, and therefore convergent;

if L is greater than 1, then the series is divergent;

if L is equal to 1, the series may be convergent, divergent, or conditionally convergent.

Now, use the above formula and evaluate the limit:

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

which simplified becomes

$\lim_{n\rightarrow \infty }\left | \frac{2}{n+3}\right |=0<1$

Because L is less than 1, the series is (absolutely) convergent.

Report an Error

Example Question #121 : Series In Calculus

Consider the series

$\sum_{n=0}^{\infty} \frac{1}{n^2+10}$ .

Using the ratio test, what can we conclude regarding its convergence?

Possible Answers:

Ratio test implies that the series is divergent.

Ratio test implies that the series is absolutely convergent.

Ratio test implies that the series is conditionally convergent.

We can conclude nothing using the ratio test. Another test will have to be used to test its convergence.

Correct answer:

We can conclude nothing using the ratio test. Another test will have to be used to test its convergence.

Explanation:

Let's use the ratio test to see what we can conclude:

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{1}{(n+1)^2+10}\frac{n^2+10}{1} \right |=1$

Since the ratio test limit results in one, we cannot conclude anything about the series' convergence/divergence by definition of the ratio test.

Report an Error

Example Question #123 : Series In Calculus

Find the radius of convergence of the following power series:

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

Possible Answers:

$\infty$

Correct answer:

$\infty$

Explanation:

Let's use the ratio test to find the radius of convergence of

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

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{x^{n+1}}{(2n+2)!}\frac{(2n)!}{x^n} \right |=\lim_{n\to \infty} \left|\frac{x}{(2n+1)(2n+2)} \right |$

independent of . This means the series converges for all , so the radius of convergence is $\infty$ .

Report an Error

Example Question #81 : Convergence And Divergence

Find the radius of convergence of the power series.

$\sum_{n=0}^{\infty} \frac{x^{7n}(2n)!}{n!n!}$

Possible Answers:

R=0

$R=4^{\frac{1}{7}}$

$R=\frac{1}{4^{\frac{1}{7}}}$

$R=\frac{1}{4^7}$

Correct answer:

$R=\frac{1}{4^{\frac{1}{7}}}$

Explanation:

We can find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{x^{7n}(2n)!}{n!n!}$ using the ratio test:

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{x^{7n+7}(2n+2)!}{(n+1)!(n+1)!}\frac{n!n!}{x^{7n}(2n)!} \right |=\lim_{n\to \infty} \left|\frac{x^7(2n+1)(2n+2)}{(n+1)(n+1)} \right |$

$=\left|4x^7 \right |<1$

So then we have

$|x|<\frac{1}{4^{\frac{1}{7}}}=R$

Which means

$R=\frac{1}{4^{\frac{1}{7}}}$

is the radius of convergence.

Report an Error

Example Question #2911 : Calculus Ii

Find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{x^{5n}}{3^n}$ .

Possible Answers:

$\infty$

$\frac{1}{3^{\frac{1}{5}}}$

$5^{\frac{1}{3}}$

$3^{\frac{1}{5}}$

Correct answer:

$3^{\frac{1}{5}}$

Explanation:

We can find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{x^{5n}}{3^n}$

using the ratio test:

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{x^{5n+5}}{3^{n+1}}\frac{3^n}{x^{5n}} \right |=\lim_{n\to \infty} \left|\frac{x^{5}}{3} \right |=\left|\frac{x^5}{3}\right|<1$

which means that $|x|<3^{\frac{1}{5}}$

So that $R=3^{\frac{1}{5}}$ is the radius of convergence.

Report an Error

Example Question #92 : Ratio Test

Determine whether the series is convergent or divergent:

$\sum_{n=0}^{\infty }\frac{n!}{3^{n+2}}$

Possible Answers:

The series may be (absolutely) convergent, divergent, or conditionally convergent

The series is divergent

The series is (absolutely) convergent

The series is conditionally convergent

Correct answer:

The series is divergent

Explanation:

To determine whether the series given converges or diverges, we must use the Ratio Test, which states that for

$L=\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_{n}}\right |$ , if L<1 , then series is (absolutely) convergent; if L=1 , then the series may be (absolutely) convergent, divergent, or conditionally convergent; and if L>1 , then the series is divergent.

So, we must evaluate the limit as given by the above formula:

$L=\lim_{n\rightarrow \infty } \frac{(n+1)!}{3^{n+3}}\cdot \frac{3^{n+2}}{(n)!}=\lim_{n\rightarrow \infty } \frac{n+1}{3}=\infty >1$

Thus, the series is divergent.

(The limit was solved by $\lim_{n\rightarrow \infty } \frac{n}{n}(\frac{1+n^{-1}}{3n^{-1}})=\infty$ . As approaches infinity, the negative terms go to zero, which makes zero in the denominator and in the numerator, equalling infinity.)

Report an Error

Example Question #93 : Ratio Test

Determine if the series converges

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

Possible Answers:

Series diverges

Cannot be determined

Series converges

Correct answer:

Series converges

Explanation:

In order to test the convergence of the series, we use the ratio test.

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

If...

L<1 the series is absolutely convergent

L=1 the convergence is undetermined

L>1 the series diverges

For this problem

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

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

$=\lim_{n\rightarrow \infty }\left | \frac{3\cdot3^{n}}{2^2\cdot2^{2n+1} } \cdot \frac{2^{2n+1}}{3^n} \right |$

$=\lim_{n\rightarrow \infty }\left | \frac{3}{2^2} } \right |$

$=\frac{3}{4}$

And since

L<1 the series absolutely converges.

Report an Error

View Calculus 2 Tutors

Yixuan
Certified Tutor

Macalester College, Bachelor in Arts, Mathematics and Statistics.

View Calculus 2 Tutors

Riley
Certified Tutor

Brigham Young University-Idaho, Bachelor of Science, Mathematics.

View Calculus 2 Tutors

Filiberto
Certified Tutor

University of Arizona, Bachelor of Science, Electrical Engineering.

All Calculus 2 Resources

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

Popular Subjects

Algebra Tutors in Houston, GRE Tutors in Dallas Fort Worth, ACT Tutors in Philadelphia, Algebra Tutors in Miami, LSAT Tutors in Washington DC, Statistics Tutors in Miami, SSAT Tutors in Dallas Fort Worth, Algebra Tutors in San Diego, Calculus Tutors in Los Angeles, Statistics Tutors in Boston

Popular Courses & Classes

ISEE Courses & Classes in Houston, SAT Courses & Classes in Atlanta, LSAT Courses & Classes in Houston, MCAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Chicago, GMAT Courses & Classes in Washington DC, ACT Courses & Classes in Houston, LSAT Courses & Classes in Los Angeles, SSAT Courses & Classes in San Diego, SAT Courses & Classes in Washington DC

Popular Test Prep

SAT Test Prep in Atlanta, ACT Test Prep in Houston, SAT Test Prep in Chicago, GMAT Test Prep in San Diego, SSAT Test Prep in Houston, ISEE Test Prep in Atlanta, MCAT Test Prep in Atlanta, SAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Seattle, ACT Test Prep in Chicago

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 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #83 : Convergence And Divergence

Example Question #2902 : Calculus Ii

Example Question #85 : Convergence And Divergence

Example Question #121 : Series In Calculus

Example Question #121 : Series In Calculus

Example Question #123 : Series In Calculus

Example Question #81 : Convergence And Divergence

Example Question #2911 : Calculus Ii

Example Question #92 : Ratio Test

Example Question #93 : Ratio Test

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: