MATH 156 University of Arizona Calculus Test Questions

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gebl27

Mathematics

Math 156

University of Arizona

MATH

Description

Attached you will find 10 Questions. Please, solve them with steps using calculus 2 methods.

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Math 156 Spring 2020 Test 3 – Part A This test has a total of 63 points, but is only worth 60 points. It is possible to get 3 points of extra credit. • Follow instructions for each problem carefully. • Show all of the steps. No credit will be given for answers without supporting work. • You must submit your own work. • Make sure that your solutions are neat and your files are readable (not faint or blurry) and oriented correctly (not sideways). 1. (8 points) Determine whether the following series is absolutely convergent, conditionally convergent, or divergent. ∞ X 1 + 4 sin(3n) √ n=1 n + 7n - Clearly state the name of the correct test(s) that you used to reach the correct conclusion. - Show all work. - Clearly state your conclusion. 2. (8 points) Determine whether the following series is absolutely convergent, conditionally convergent, or divergent. ∞ X 5k 2 + 1 √ 3k 8 − 2k k=1 - Clearly state the name of the correct test(s) that you used to reach the correct conclusion. - Show all work. - Clearly state your conclusion. 3. (8 points) Determine whether the following series is absolutely convergent, conditionally convergent, or divergent. ∞ X (−1)n+1 n=1 (3n − 1)2n 2n (3n2 + 1)n - Clearly state the name of the correct test(s) that you used to reach the correct conclusion. - Show all work. - Clearly state your conclusion. 4. (8 points) Determine whether the following series is absolutely convergent, conditionally convergent, or divergent. ∞ X (−1)n−1 n=1 5n − 2 - Clearly state the name of the correct test(s) that you used to reach the correct conclusion. - Show all work. - Clearly state your conclusion. 5. (6 points) Determine whether each of the following series is convergent or divergent. Justify your answers. Be specific. ∞ X (a) √ n=1 ∞ X n−1 3n5 + 1 5n + 1 ln (b) 3n − 1 n=1 ∞  X (c)   3−k + 3k −3  k=1 6. (6 points) Give examples of the following. Explain why your example satisfies required conditions. Make sure that your examples are simple. (a) An alternating series that is absolutely convergent. (b) An alternating series that is conditionally convergent. (c) An alternating series that is divergent. 7. (8 points) Find the center and the radius of convergence for the following series. Show all work (all algebraic steps). ∞ X (−1)n n=0 n2 (x + 5)n 7n 8. (5 points) The power series ∞ X (−1)n n=1 (x − 1)n √ has the radius of convergence R = 3. 3n 3 n Find the interval of convergence for this series. - Show all work (all algebraic steps). - Represent your answer in interval notation. 9. (3 points) Determine whether the interval [−5, 3) can represent an interval of convergence for a X power series cn (x + 2)n . Justify your answer. 10. Bonus: (3 points) Let X cn (x − 2)n be a power series that is conditionally convergent when x = 7. Determine whether the series P cn 6n is convergent or divergent. Justify your answer. Note: coefficients cn are not known, but are the same in both series.
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Explanation & Answer

Well, please read the solutions and tell me if something is unclear.

Estimate the general term:
|

5

The series ∑ 7𝑛

1 + 4 sin(3𝑛)

|≤

1+4
5
=
.
7𝑛
7𝑛

√𝑛 + 7𝑛
is (absolutely) convergent, so we can apply the direct comparison test

and state that the initial series converges absolutely.

Again, estimate the general term:
5𝑘 2 + 1
5𝑘 2 + 𝑘 2
6𝑘 2
6
|
|≤
= 4 = 2.
𝑘
𝑘
√3𝑘 8 − 2𝑘
√3𝑘 8 − 2𝑘 8
6

The series ∑ 𝑘 2 is (absolutely) convergent, so we can apply the direct comparison test
and state that the initial series converges absolutely.

Apply the root test:
(3𝑛 − 1)2
9𝑛2 − 6𝑛 + 1 9 3
√|𝑎𝑛 | =
=
→ = > 1.
2(3𝑛2 + 1)
6𝑛2 + 2
6 2

𝑛

Because the limit exists and is greater than 1, so the limit superior. Therefore, the series is
divergent.

It is an alternating series with the absolute value of the general term monotone tending to
zero:
1
𝑎𝑛 = (−1)𝑛+1 𝑏𝑛 , 𝑏𝑛 =
→ 0 and 𝑏𝑛+1 < 𝑏𝑛 .
5𝑛 − 2
Thus, the series converges by the alternating series test.
1

It does not converge absolutely because |𝑎𝑛 | = 𝑏𝑛 ≥ 5𝑛, which is divergent (apply the
direct comparison test here).
The answer is: the series is conditionally convergent.

Estimate
|
3

2

𝑛−1
√3𝑛5 + 1

|≤

𝑛
√3𝑛5

=

1
√3

3

𝑛 −2 .

The series ∑ 𝑛 is convergent, therefore the given series is also (absolutely) convergent
by the direct comparison test.

5

The limit of the general term is obviously ln (3), which is nonzero. Therefore...


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