AP Calculus BC : Alternating Series with Error Bound

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

Example Question #71 : Polynomial Approximations And Series

Determine whether the following series converges or diverges:

Possible Answers:

The series may (absolutely) converge, diverge, or conditionally converge

The series diverges

The series (absolutely) converges

The series conditionally converges

Correct answer:

The series (absolutely) converges


Given just the harmonic series, we would state that the series diverges. However, we are given the alternating harmonic series. To determine whether this series will converge or diverge, we must use the Alternating Series test. 

The test states that for a given series where  or  where  for all n, if  and  is a decreasing sequence, then  is convergent.

First, we must evaluate the limit of  as n approaches infinity:

The limit equals zero because the numerator of the fraction equals zero as n approaches infinity. 

Next, we must determine if  is a decreasing sequence. , thus the sequence is decreasing.

Because both parts of the test passed, the series is (absolutely) convergent.

Example Question #1 : Alternating Series With Error Bound

Determine whether 

converges or diverges, and explain why.

Possible Answers:

Divergent, by the comparison test.

Convergent, by the alternating series test. 

Divergent, by the test for divergence.

Convergent, by the -series test.

More tests are needed.

Correct answer:

Convergent, by the alternating series test. 


We can use the alternating series test to show that


We must have   for  in order to use this test. This is easy to see because  is in for all  (the values of this sequence are ), and sine is always nonzero whenever sine's argument is in .

Now we must show that


2.  is a decreasing sequence.

The limit 

implies that 

so the first condition is satisfied.

We can show that  is decreasing by taking its derivative and showing that it is less than  for :

The derivative is less than , because  is always less than , and that  is positive for , using a similar argument we used to prove that  for . Since the derivative is less than  is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that 

converges, by the alternating series test.

Example Question #2 : Alternating Series With Error Bound

For the series:  , determine if the series converge or diverge.  If it diverges, choose the best reason.

Possible Answers:

Correct answer:


The series given is an alternating series.  

Write the three rules that are used to satisfy convergence in an alternating series test.

For :

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since  and instead approaches infinity.

The correct answer is:

Example Question #1 : Alternating Series With Error Bound

Determine whether the series converges or diverges:

Possible Answers:

The series is conditionally convergent.

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

The series is (absolutely) convergent.

The series is divergent.

Correct answer:

The series is divergent.


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

 - and  where  for all n - to converge, 

 must equal zero and  must be a decreasing series.

For our series, 


because it behaves like 


The test fails because  so we do not need to check the second condition of the test.

The series is divergent.



Example Question #2 : Lagrange Error

Which of the following series does not converge?

Possible Answers:

Correct answer:


We can show that the series   diverges using the ratio test.




 will dominate over  since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence. 

Alternatively, it's clear that  is much greater than , and thus having  in the numerator will make the series diverge by the  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.




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