We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus BC : Series of Constants

Study concepts, example questions & explanations for AP Calculus BC

Create An Account Create Tests & Flashcards

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 Next →

Example Question #4 :Alternating Series With Error Bound

Determine whether the series converges or diverges:

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

Possible Answers:

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

The series is (absolutely) convergent.

The series is conditionally convergent.

The series is divergent.

有限公司rrect answer:

The series is divergent.

Explanation:

To determine whether the series converges or diverges, we must use the Alternating Series test, which states that for

$\sum a_{n}$ - and $a_{n}=(-1)^{n+1}b_{n}$ where $b_{n} \geq 0$ for all n - to converge,

$\lim_{n\rightarrow \infty }b_{n}$ must equal zero and $b_{n}$ must be a decreasing series.

For our series,

$\lim_{n\rightarrow \infty }b_{n}=\lim_{n\rightarrow \infty }\frac{n^4}{2n^4+1}=\frac{1}{2}$

because it behaves like

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

The test fails because $\frac{1}{2}\neq0$ so we do not need to check the second condition of the test.

The series is divergent.

Report an Error

Example Question #1 :Abstract Algebra

Which of the following series does not converge?

Possible Answers:

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

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

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

$\sum_{i=0}^{\infty} 0.9999999999999^n$

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

有限公司rrect answer:

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

Explanation:

We can show that the series $\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ diverges using the ratio test.

$L = \lim_{n ->\infty } \frac{a_{n+1}}{a_{n}} = \lim_{n ->\infty } \left[\left(\frac{(n+1)!}{(n+1)^2 cos (n+1)} \div \frac{n!}{n^2 cos (n) }\right)\right]$

$= \lim_{n ->\infty} \frac{n^2(n+1)cos(n)}{(n+1)^2cos(n+1)} = \lim_{n ->\infty} \frac{n^2cos(n)}{(n+1)cos(n+1)}$

n^2 will dominate over (n+1) since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear thatis much greater than n^2 , and thus havingin the numerator will make the series diverge by the $n^{th}$ limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

Report an Error

Example Question #1 :有限公司ncepts Of Convergence And Divergence

One of the following infinite series CONVERGES. Which is it?

Possible Answers:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$

$\sum_{k=1}^{\infty} (-1)^k sin(k)$

$\sum_{k=2}^{\infty} \frac{x}{ln(x)}$

None of the others converge.

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

有限公司rrect answer:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$

Explanation:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ converges due to the comparison test.

We start with the equation $\frac{1}{k^2} =\frac{1}{k^2}$ . Since $sin(k) \leq 1$ for all values of k, we can multiply both side of the equation by the inequality and get $\frac{sin(k)}{k^2} \leq \frac{1}{k^2}$ for all values of k. Since $\sum_{k=1}^{\infty} \frac{1}{k^2}$ is a convergent p-series with p=2 , $\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ hence also converges by the comparison test.

Report an Error

Example Question #1 :P Series

Determine the nature of convergence of the series having the general term:

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

Possible Answers:

The series is convergent.

$\frac{\pi}{5}$

$\frac{\pi}{7}$

The series is divergent.

$\frac{1}{2}$

有限公司rrect answer:

The series is convergent.

Explanation:

We will use the Limit Comparison Test to establish this result.

We need to note that the following limit

$\frac{\frac{8}{\sqrt{n(n+1)(n+2)}}}{\frac{8}{n^{3/2}}}}$

goes to 1 as n goes to infinity.

Therefore the series have the same nature. They either converge or diverge at the same time.

We will focus on the series:

$\frac{1}{n^{3/2}}$ .

We know that this series is convergent because it is a p-series. (Remember that

$\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges if p>1 and we have p=3/2 which is greater that one in this case)

By the Limit Comparison Test, we deduce that the series is convergent, and that is what we needed to show.

Report an Error

Example Question #1 :P Series

Determine if the series converges or diverges. You do not need to find the sum.

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

Possible Answers:

Diverges

There is not enough information to decide convergence.

Neither converges nor diverges.

有限公司nditionally converges.

有限公司nverges

有限公司rrect answer:

有限公司nverges

Explanation:

We can compare this to the series $\sum_{n = 1}^{\infty } \frac{1}{n^{3}}$ which we know converges by the p-series test.

To figure this out, let's first compare $n^{3} + 2$ to $n^{3}$ . For any number n, $n^{3} + 2$ will be larger than $n^{3}$ .

There is a rule in math that if you take the reciprocal of each term in an inequality, you are allowed to flip the signs.

Thus, $n^{3} + 2 > n^{3}}$ turns into

$\frac{1}{n^{3}} > \frac{1}{n^{3} + 2}$ .

And so, because $\frac{1}{n ^{3}}$ converges, thus our series also converges.

Report an Error

Example Question #1 :Harmonic Series

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

Possible Answers:

Integral Test: The improper integral determines that the harmonic series diverge.

$\textup{}$ Nth Term Test: The series diverge because the limit asgoes to infinity is zero.

P-Series Test: The summation converges since p=1 .

$\textup{}$ Divergence Test: Since limit of the series approaches zero, the series must converge.

Root Test: Since the limit asapproaches to infinity is zero, the series is convergent.

有限公司rrect answer:

Integral Test: The improper integral determines that the harmonic series diverge.

Explanation:

The series $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series.

The Nth term test and the Divergent test may not be used to determine whether this series converges, since this is a special case. The root test also does not apply in this scenario.

According the the P-series Test, $\sum_{x=1}^{\infty}\frac{1}{x^p}$ must converge only if p>1 . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for $p \leq1$ .

This leaves us with the Integral Test.

$\int_{1}^{\infty}\frac{1}{x}= \lim_{b \to \infty}\int_{1}^{b}\frac{1}{x}dx=\lim_{b \to \infty}( ln(b)-ln(1))=\infty$

由于反常积分发散,那么series.

Report an Error

Example Question #1 :Alternating Series

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

Possible Answers:

Cannot tell with the given information.

Does not exist.

有限公司nverge Absolutely.

Diverges.

有限公司nverge Conditionally.

有限公司rrect answer:

有限公司nverge 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.

Report an Error

Example Question #11 :Series Of Constants

True or False, a-series cannot be tested conclusively using the ratio test.

Possible Answers:

True

False

有限公司rrect answer:

True

Explanation:

We cannot test for convergence of a-series using the ratio test. Observe,

For the series $\sum_{n=1}^\infty \frac{1}{n^p}$ ,

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} | \frac{n^p}{(n+1)^p}| = \lim_{n \to \infty} (\frac{n}{n+1})^p = 1^p =1$ .

Since this limit isregardless of the value for, the ratio test is inconclusive.

Report an Error

Example Question #1 :Ratio Test And Comparing Series

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

$\sum_{n=1}^{\infty} \frac{(-9)^n}{5^{3n+2}(n+1)}$

Possible Answers:

Neither

Inconclusive

有限公司nvergent

Divergent

Both

有限公司rrect answer:

有限公司nvergent

Explanation:

In order to figure out if

$\sum_{n=1}^{\infty} \frac{(-9)^n}{5^{3n+2}(n+1)}$

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

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

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

Then if

L<1 , the series is absolutely convergent.

L>1 , the series is divergent.

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

Now lets apply the ratio test to our problem.

Let

$a_n=\frac{(-9)^n}{5^{3n+2}(n+1)}$

and

$a_{n+1}=\frac{(-9)^{n+1}}{5^{3(n+1)+2}((n+1)+1)}$

$=\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}}{\frac{(-9)^n}{5^{3n+2}(n+1)}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{(-9)^{n+1}5^{3n+2}(n+1)}{(-9)^n5^{3n+5}(n+2)} \right |$

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{(-9)^{1}(n+1)}{5^{3}(n+2)} \right |$

$=\frac{9}{125}\lim_{n\rightarrow \infty}\left | \frac{n+1}{n+2} \right |$

$=\frac{9}{125}*1=\frac{9}{125}$ .

Since

$\frac{9}{125}<1$ .

We have sufficient evidence to conclude that the series is convergent.

Report an Error

Example Question #1 :Limits Of Sequences

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

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

Possible Answers:

Both

Inconclusive

Neither

Divergent

有限公司nvergent

有限公司rrect answer:

Divergent

Explanation:

In order to figure if

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

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

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

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

Then if

L<1 , the series is absolutely convergent.

L>1 , the series is divergent.

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

Now lets apply the ratio test to our problem.

Let

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

and

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

Now

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

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

Now lets simplify this expression to

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

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

$=\frac{1}{2}*\infty=\infty$ .

Since $\infty>1$ ,

we have sufficient evidence to conclude that the series is divergent.

Report an Error

← Previous 1 2 3 4 5 Next →

有限公司pyright Notice

View AP Calculus BC Tutors

Jabin
Certified Tutor

University of Arizona, Bachelor of Science, Mathematics.

View AP Calculus BC Tutors

Angela
Certified Tutor

Saint Josephs College, Bachelor in Arts, Mathematics.

View AP Calculus BC Tutors

Nicholas
Certified Tutor

Stony Brook University, Bachelor of Science, Mechanical Engineering.

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

LSAT Tutors in San Diego,SAT Tutors in Houston,ACT Tutors in Miami,GMAT Tutors in New York City,有限公司mputer Science Tutors in Phoenix,Reading Tutors in Dallas Fort Worth,Statistics Tutors in Boston,MCAT Tutors in Atlanta,English Tutors in Atlanta,MCAT Tutors in New York City

Popular Courses & Classes

SAT Courses & Classes in Atlanta,GMAT Courses & Classes in New York City,GRE Courses & Classes in Houston,ISEE Courses & Classes in Chicago,SSAT Courses & Classes in Chicago,GMAT Courses & Classes in Seattle,ISEE Courses & Classes in San Diego,ISEE Courses & Classes in Atlanta,考试课程及课程在西雅图,SSAT Courses & Classes in Los Angeles

Popular Test Prep

ACT Test Prep in Boston,ACT Test Prep in Los Angeles,MCAT Test Prep in Boston,SSAT Test Prep in Washington DC,GMAT Test Prep in Miami,LSAT Test Prep in Denver,SAT Test Prep in Philadelphia,SSAT Test Prep in Los Angeles,GRE Test Prep in Philadelphia,GRE 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

有限公司ntact Information

*Full legal name

有限公司mpany name

*Phone number

*Email address

Address

*Address1

Address2

*City

*State

*Zipcode

*有限公司untry

有限公司mplaint Details

有限公司pyright holder you represent (if other than yourself)

*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

AP Calculus BC : Series of Constants

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #4 :Alternating Series With Error Bound

Example Question #1 :Abstract Algebra

Example Question #1 :有限公司ncepts Of Convergence And Divergence

Example Question #1 :P Series

Example Question #1 :P Series

Example Question #1 :Harmonic Series

Example Question #1 :Alternating Series

Example Question #11 :Series Of Constants

Example Question #1 :Ratio Test And Comparing Series

Example Question #1 :Limits Of Sequences

All AP Calculus BC Resources

Report an issue with this question

DMCA Complaint

有限公司ntact Information

Address

有限公司mplaint Details

Find the Best Tutors

Email address:
Your name:
Feedback: