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 #131 : Series In Calculus

Use the ratio test to figure out if the series

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

converges.

Possible Answers:

It converges

it diverges to infinity

Correct answer:

It converges

Explanation:

an easy sequence to compare $\frac{1}{n^2+n}$ to is $\frac{1}{n^2}$ , which we know converges.

$\frac{\frac{1}{n^2}}{\frac{1}{n^2+n}} = \frac{n^2+n}{n^2}$

We need to find $\lim_{n \to \infty} \frac{n^2+n}{n^2} = \lim_{n \to \infty} 1+\frac{1}{n}$

$\frac{1}{n}$ goes to zero so this limit is . Since the limit of the ratio is 1 and $\sum \frac{1}{n^2}$ converges, $\sum \frac{1}{n^2+n}$ must converge also.

Report an Error

Example Question #92 : Ratio Test

Can we use the Ratio Test to test the convergence/divergence of the infinite series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3}$ ?

Possible Answers:

No, the ratio test fails here.

Yes, and the series converges

Yes, and the series diverges

Correct answer:

No, the ratio test fails here.

Explanation:

To use the ratio test, we must first evaluate

$\lim_{k \to \infty} |\frac{a_{k+1}}{a_k}|$

$=\lim_{k \to \infty} |\frac{\frac{(-1)^{k+1}}{(k+1)^3}}{\frac{(-1)^k}{k^3}}|$

$=\lim_{k \to \infty} |\frac{(-1)^{k+1}}{(k+1)^{3}} \times \frac{k^3}{(-1)^k}|$

$= \lim_{k \to \infty} | -1 \times \frac{k^3}{(k+1)^3}|$

$= \lim_{k \to \infty} \frac{k^3}{(k+1)^3}$

Since the result of the limit is , the Ratio Test fails, and therefore we cannot use it in testing this series for convergence/divergence.

Report an Error

Example Question #132 : Series In Calculus

Determine whether the series converges or diverges:

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

Possible Answers:

The series may be convergent, divergent, or conditionally convergent

The series diverges

The series is conditionally convergent

The series converges

Correct answer:

The series diverges

Explanation:

To determine whether the series converges or diverges, we must use the ratio test, which states that for the given series $\sum_{n=0}^{\infty} a_{n}$ , and

$L=\lim_{n\rightarrow \infty} \left |\frac{ a_{n+1}}{a_{n}}\right |$ , when L is equal to 1, then the series may be convergent, divergent, or conditionally convergent; when L is less than 1 then the series is convergent; and when L is greater than 1 the series is divergent.

Using the above test, we get

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

Because L is greater than 1, the series diverges.

Report an Error

Example Question #96 : Ratio Test

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

Determine the convergence of the series using the Ratio Test.

Possible Answers:

Series diverges

Cannot be determined

Series converges

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |$

where a_n is the n-th term of the series.

if the limit

is greater than , the series diverges
is less than , the series converges
equal to , the test is inconclusive

Finding the limit for the terms in our series,

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)^2}{3^{n+1}}}{\frac{n^2}{3^n}} \right |$

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

$=\frac{1}{3}\lim_{n \to \infty}\left | \frac{n^2+2n+1}{n^2} \right |$

$=\frac{1}{3}*1$

$=\frac{1}{3}$

Because the limit is less than ,

the series converges.

Report an Error

Example Question #97 : Ratio Test

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

Determine the convergence of the series using the Ratio Test.

Possible Answers:

Series converges

Cannot be determined

Series diverges

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |$

where a_n is the n-th term of the series.

if the limit

is greater than , the series diverges
is less than , the series converges
equal to , the test is inconclusive

Finding the limit for the terms in our series,

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(ln\,n+1)3^{n+1}}{2^{2(n+1)+1}}}{\frac{(ln\,n)3^n}{2^(2n+1)}} \right |$

$=\lim_{n \to \infty}\left |\frac{(ln\,n+1)2^{2n+1}*3^{n+1}}{(ln\,n)2^{2n+3}*3^n} \right |$

$=\frac{3}{2^2}\lim_{n \to \infty}\left |\frac{ln\,n+1}{ln\,n} \right |$

$=\frac{3}{4}*1$

$=\frac{3}{4}$

Because the limit is less than ,

the series converges.

Report an Error

Example Question #93 : Ratio Test

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

Possible Answers:

Series converges

Cannot be determined

Series diverges

Correct answer:

Series diverges

Explanation:

The ratio test is defined as follows

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |$

where a_n is the n-th term of the series.

if the limit

is greater than , the series diverges
is less than , the series converges
equal to , the test is inconclusive

Finding the limit for the terms in our series,

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)!}{4^{n+1}}}{\frac{n!}{4^n}} \right |$

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

$=\frac{1}{4}\lim_{n \to \infty}\left |n+1\right |$

$=\frac{1}{4}*\infty$

$=\infty$

Because the limit is greater than ,

the series diverges.

Report an Error

Example Question #98 : Ratio Test

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

Determine the convergence of the series.

Possible Answers:

Series converges

Cannot be determined

Series diverges

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |$

where a_n is the n-th term of the series.

if the limit

is greater than , the series diverges
is less than , the series converges
equal to , the test is inconclusive

Finding the limit for the terms in our series,

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

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

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

$=\frac{1}{2}*0$

Because the limit is less than ,

the series converges.

Report an Error

Example Question #101 : Ratio Test

Determine the convergence or divergence of the following series:

$\sum_{n=0}^{\infty}\frac{(n+3)!}{10^{n+1}}$

Possible Answers:

The series is divergent

The series is convergent

The series is conditionally convergent

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

Correct answer:

The series is divergent

Explanation:

To determine the convergence or divergence of the following series, we must use the Ratio Test, which states that for the series $\sum_{n=0}^{\infty}a_{n}$ , and $L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_{n}} \right |$ , that when L>1 , the series diverges, when L<1 , the series converges, and when L=1 , the series may be absolutely convergent, conditionally convergent, or divergent.

For our series, using the above formula, we get $L=\lim_{n\rightarrow \infty} \left | \frac{(n+4)!}{10^{n+2}}\cdot \frac{10^{n+1}}{(n+3)!}\right |$

which simplified becomes

$L=\lim_{n\rightarrow \infty} \left | \frac{n+4}{10}\right |=\infty$

Note that the factorial was simplified by rewriting (n+4)!=(n+4)(n+3)! .

In the limit, the n in the numerator goes to infinity, which makes L go to infinity, which is greater than 1. Thus, the series diverges.

Report an Error

Example Question #102 : Ratio Test

Determine if the series converges or diverges:

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

Possible Answers:

The series diverges

The series converges

The series may absolutely converge, conditionally converge, or diverge

The series conditionally converges

Correct answer:

The series converges

Explanation:

To determine the convergence or divergence of the series, we must use the ratio test, which states that for a given series $\sum a_{n}$ , and $L=\lim_{n\rightarrow \infty} \left | \frac{a_{n+1}}{a_n}\right |$ , if L is greater than 1, the series diverges, if less than 1, the series converges, and if L is equal to 1, the series may absolutely converge, conditionally converge, or diverge.

Now, we must find L:

$L=\lim_{n\rightarrow \infty} \left | \frac{2^{n+3}}{(n+3)!} \cdot \frac{(n+2)!}{2^{n+2}}\right |=\lim_{n\rightarrow \infty}\left | \frac{2}{n+3} \right |=0$

We simplified the limit using the properties of radicals and exponents. The denominator of the limit becomes infinitely large, so the term goes to zero. L is therefore less than 1, so the series converges.

Report an Error

Example Question #133 : Series In Calculus

Determine if the series converges (absolutely or conditionally) or diverges:

$\sum_{n=0}^{\infty}\frac{n^2+1}{5^n}$

Possible Answers:

Absolutely Converges

May Absolutely or Conditionally converge, or diverge.

Diverges

Conditionally Converges

Correct answer:

Absolutely Converges

Explanation:

Using the ratio test $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}$ , we get $\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2+1}{5^{n+1}}*\frac{5^n}{n^2+1}\right |=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2+1}{5n^2+5} \right |$ . Evaluating the limit, we get $\frac{1}{5}$ . Because $\frac{1}{5}<1$ the series converges. Also, because no n value could make the series negative, it absolutely converges .

Report an Error

View Calculus 2 Tutors

Mayowa
Certified Tutor

Logan University, Bachelor of Science, Biology, General.

View Calculus 2 Tutors

Leonid
Certified Tutor

Polytechnic Institute of Bucharest, Romania, Bachelor of Science, Avionics Maintenance Technology. Bucharest Academy of Econo...

View Calculus 2 Tutors

Aaron
Certified Tutor

Tulsa Community College, Associate in Science, Mechanical Engineering.

All Calculus 2 Resources

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

Popular Subjects

Computer Science Tutors in Houston, Reading Tutors in Philadelphia, English Tutors in New York City, GRE Tutors in Denver, MCAT Tutors in Chicago, Math Tutors in Miami, Spanish Tutors in Philadelphia, Physics Tutors in Philadelphia, GRE Tutors in New York City, Calculus Tutors in San Diego

Popular Courses & Classes

LSAT Courses & Classes in Miami, GMAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Miami, LSAT Courses & Classes in Chicago, GRE Courses & Classes in Phoenix, SAT Courses & Classes in Denver, MCAT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Los Angeles, Spanish Courses & Classes in Atlanta, SSAT Courses & Classes in Houston

Popular Test Prep

LSAT Test Prep in Miami, GMAT Test Prep in New York City, SAT Test Prep in Denver, GMAT Test Prep in Washington DC, ACT Test Prep in Seattle, ISEE Test Prep in San Diego, ISEE Test Prep in Chicago, ISEE Test Prep in New York City, SSAT Test Prep in San Diego, SAT Test Prep in New York City

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 #131 : Series In Calculus

Example Question #92 : Ratio Test

Example Question #132 : Series In Calculus

Example Question #96 : Ratio Test

Example Question #97 : Ratio Test

Example Question #93 : Ratio Test

Example Question #98 : Ratio Test

Example Question #101 : Ratio Test

Example Question #102 : Ratio Test

Example Question #133 : 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: