Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In Exercises 52–57, do each of the following:

(a) Show that the given alternating series converges.

(b) Compute S10 and use Theorem 7.38 to find an interval containing the sum L of the series.

(c) Find the smallest value of n such that Theorem 7.38 guarantees that Sn is within 106 of L.

k=0(1)k(2k+1)!

Short Answer

Expert verified

(a). Using the test for convergence for alternating series we can see that the given alternating series converges.

(b). The value of S10 is 113!+15!17!+19!111!+113!115!+117!119!+121!0.841470984

and the value of L is L(0.84147098480789650665254,0.8414709848078965066526).

Step by step solution

01

Step 1. Given Information

Given, an alternating series

k=0(1)k(2k+1)!

02

Step 2. Using the alternating series test

Alternating series test for convergence is stated as if {ak} be a strictly decreasing sequence of positive numbers such that limkak=0 . Then the alternating series k=1(1)kak and k=1(1)k+1ak both converges.

03

Step 3. Applying the alternating series test to the given series 

The series k=01(2k+1)! is clearly a nonnegative strictly decreasing series. Also,

limk1(2k+1)!=0. Hence, by alternating series test this series converges and the series k=0(1)k(2k+1)! also converges.

04

Step 4. Given information

We are given with Theorem 7.8 to approximate the partial sum S10

05

Step 5. Calculating S10

If we add the first 10 terms of the series to obtain the 10th partial sum, S10, we obtain

S10=113!+15!17!+19!111!+113!115!+117!119!+121!0.841470984

This is an approximation for the sum L of the series. By Theorem 7.38, the error in using S10 is at most 123!3.86817×1023. In addition, Theorem 7.38 tells us that S10 is less than the sum of the series, L, because LS10has the same sign as the term 123!3.86817×1023 . Also,

LS10<a113.86817×1023. We have, L(S10,S10+a11). Therefore, L(0.84147098480789650665254,0.8414709848078965066526)

06

Step 6. Determining n

We have to find n such that

Sn<106

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine whether the series k=2-35kconverges or diverges. Give the sum of the convergent series.

Let αbe any real number. Show that there is a rearrangement of the terms of the alternating harmonic series that converges to α. (Hint: Argue that if you add up some finite number of the terms of k=112k1, the sum will be greater than α. Then argue that, by adding in some other finite number of the terms of

k=112k , you can get the sum to be less than α. By alternately adding terms from these two divergent series as described in the preceding two steps, explain why the sequence of partial sums you are constructing will converge to α.)

Leila, in her capacity as a population biologist in Idaho, is trying to figure out how many salmon a local hatchery should release annually in order to revitalize the fishery. She knows that ifpksalmon spawn in Redfish Lake in a given year, then only 0.2pkfish will return to the lake from the offspring of that run, because of all the dams on the rivers between the sea and the lake. Thus, if she adds the spawn from h fish, from a hatchery, then the number of fish that return from that run k will be pk+1=0.2(pk+h)..

(a) Show that the sustained number of fish returning approaches p=hk+10.2kas k→∞.

(b) Evaluate p.

(c) How should Leila choose h, the number of hatchery fish to raise in order to hold the number of fish returning in each run at some constant P?

Let 0 < p < 1. Evaluate the limitlimk1/klnk1/kp

Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the seriesk=21klogk

Find the values of x for which the seriesk=0xkconverges.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free