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

Do the following to k=1-1kk2:

Part (a): Show that the given alternating series converges.

Part (b): Compute S10and use Theorem 7.38to find an interval containing the sum L of the series.

Part (c): Find the smallest value of n such that Theorem 7.38guarantees that Snis within 10-6 of L.

Short Answer

Expert verified

Part (a): To show that the given alternating series converges, use the conditions of alternating series test.

Part (b): The interval containing the sum L of the series is 0.08179,0.08262.

Part (c): The smallest value of n is999.

Step by step solution

01

Part (a) Step 1. Given information.

Consider the given question,

k=1-1kk2

02

Part (a) Step 2. To show the alternating series converges.

The sequence ak=1k2is a sequence of positive numbers and is a decreasing sequence as 1k+12<1k2.

Also, limk1k2=0

Thus, the conditions of alternating series test are fulfilled.

Hence, the alternating series k=1-1kk2 is convergent.

03

Part (b) Step 1. Find the interval containing the sum L of the series.

The value S10is obtained by adding first 10terms of the series to obtain the 10thpartial sum. Therefore,S10=1-122+132-...-1102=0.8179

Thus, the value ofS10is0.8179.

04

Part (b) Step 2.  Assume L to be the sum of an alternating series.

Assume L to be the sum of an alternating series satisfying the hypothesis of the alternating series test.

For any term Snis the sequence of partial sums, L-Sn<an+1

Furthermore, the sign of the difference L-Snis the sign of the coefficient of the term an+1.

The error in using S10is at almost 1112=0.0083. That is given below,

L-0.8179<0.0083

The approximation S10 is less than L as the coefficient of a11is positive.

Thus, actual value of L lies in the interval S10,S10+a11.

Then, the interval is S10,S10+a11=0.08179,0.08179+0.0083=0.08179,0.08262

05

Part (c) Step 1. Find the smallest value of n.

The smallest value of n is given below,

1n+1210-6n+1-210-61000n+1999n

Thus, the smallest value of n is999.

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

Study anywhere. Anytime. Across all devices.

Sign-up for free