Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Q. 59 (page 592)

In Exercises 59–62 use the derivative test in Theorem 7.6 to analyze the monotonicity of the given sequence.

k+1k

Short Answer

Expert verified

The given sequence is strictly decreasing.

Step by step solution

01

Step 1. Given Information.

The given sequence isk+1k.

02

Step 2. Use the derivative test.

To analyze the monotonicity of the given sequence we will use the derivative test.

Let the function is f(k)=k+1k.

According to the derivative test,

f'(k)=k12k+1-k+1k2f'(k)=k-2k+12k+1k2f'(k)=k-2k-2k22k+1f'(k)=-k-2k22k+1f'(k)=-k+2k22k+1

Now, f'k<0forallk>0.

Therefore, the given sequence is strictly decreasing.

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.

Start your free trial

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

For each series in Exercises 44–47, do each of the following:

(a) Use the integral test to show that the series converges.

(b) Use the 10th term in the sequence of partial sums to approximate the sum of the series.

(c) Use Theorem 7.31 to find a bound on the tenth remainder R10.

(d) Use your answers from parts (b) and (c) to find an interval containing the sum of the series.

(e) Find the smallest value of n so that.

∑k=2∞1k(lnk)2

Prove Theorem 7.31. That is, show that if a function a is continuous, positive, and decreasing, and if the improper integral ∫1∞a(x)dxconverges, then the nth remainder, Rn, for the series∑k=1∞a(k) is bounded by0≤Rn=∑k=n+1∞a(k)≤∫n∞a(x)dx

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=1∞ 12k−1, the sum will be greater than α. Then argue that, by adding in some other finite number of the terms of

∑k=1∞ −12k , 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.)

Use either the divergence test or the integral test to determine whether the series in Given Exercises converge or diverge. Explain why the series meets the hypotheses of the test you select.

∑k=1∞ tan−1⁡k1+k2

Determine whether the series 9940-9920+9910-995+…converges or diverges. Give the sum of the convergent series.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation
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