Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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- (2) An interesting fact is that the Root Test (stated at the end of the previous problem) is stronger than the Ratio Test. Proving this fact would require showing two things: First, one can show that any time the Ratio Test limit exists and equals some L then the Root Test limit also exists and equals the same L. Second, one can find examples of series where the Root Test works but the Ratio Test does not. Your task in this problem is to do the second thing above. To be more specific, find an example of a series a, such that • lim Vlan[ exists and equals a number less than 1, meaning the series converges absolutely by the Root Test, and • lim does not exist. Give a brief explanation why your example satisfics the above two properties.arrow_forwardI have done a bunch of these of problems, but this one step looks wrong in the solution for this problem. https://www.bartleby.com/solution-answer/chapter-62-problem-71e-calculus-volume-2-17th-edition/9781938168062/in-the-following-exercises-express-each-series-as-a-rational-function-71-n11-x3-2n1/b209f360-2097-11e9-8385-02ee952b546e On the very first step when I distribute the (x-3)^(2n-1) exponent in the denominator to be (x-3)^2n and (x-3)^-1, the (x-3)^-1, should flip up to the numerator (not stay in the denominator as (x-3)), but it doesn't according to the answer key. And the answer is right but that first step LOOKS wrong. Can you help explain that exponent distribution?arrow_forwardQuestion 2 Use the Alternating Series Estimation Theorem to find the minimum number of terms of the infinite (-1)" we need to add to approximate the sum of the series with Jerror| < .008. series n3arrow_forward
- 4. (a) How many terms of the given alternating series do we need to add in order to find the sum of the series correct to 5 decimal places? (That is, Jerror|< 0.000005.) Show your work! (b) Approximate the sum of the series correct to 5 decimal places. Show your work! 7 (-1)7+1. 2" (n!) n=1arrow_forward12) There is one special “family” of series for which, if the series converges, we can determine the exact value of the series. This is the family where it really matters that we know the bounds of the summation. What family is this?arrow_forwardFor problem 2 part 2: To get your comparison series for LCT, choose the series with leading term over leading term. So you want the series of sqrt{n} over n "leading term over leading term" then go from therearrow_forward
- 1. In this problem, you'll look at the series > 1 )J: One can visualize this series as area, by drawing a k5/4 * k=10 box of area 103/4, a box of area 134 and so on. 115/4 (a) There are two ways to draw the series (as boxes) in question. Draw each of them together with the graph y= 34 on two different graphs. x5/4 (b) Use the picture to write an inequality comparing >) 1 to an improper integral. Check the k5/4 k=10 bounds of your improper integral carefully, and be sure you are comparing on the same domain. (c) Is the improper integral you wrote in (b) convergent or divergent? Explain. (d) Using the inequality you wrote in (b) and your answer to (c), what (if anything) can you conclude 1 :? about the convergence of 2 k5/4 k=10 (e) Repeat (b)-(d) with your second picture.arrow_forwardHello, Can someone show to solve problem 4 using the root test? Thank you!arrow_forwardI was looking for some help with this question, thanks.arrow_forward
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