Doubtrix Logo
  • home
  • Study help
    • Ask Your Doubt
  • Tutorials
  • For Tutors
  • Contact Us
  • Login
  • Sign Up
Search
Sign in | Sign Up
Search
Doubtrix Logo
  • home
  • Study help
    • Ask Your Doubt
  • Tutorials
  • For Tutors
  • Contact Us

Search questions

Subject:

Answer Type:

  • Math Archive: Questions from 2024-03-4

    Estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series ∑ n = 1 ∞ (−1) n + 1 (0.05) n n. The magnitude of the error is approximately ◻. (Simplify your answer. Use scientific notation. Use the multiplication symbol in the math palette as needed. Round to three decimal places as needed.)

    1 answer SHARE

    Estimate the magnitude of the error involved in using the sum of the first three terms to approximate the sum of the entire series ∑ n = 1 ∞ (−1) n + 1 (0.04) n n. The magnitude of the error is approximately ◻. (Simplify your answer. Use scientific notation. Use the multiplication symbol in the math palette as needed. Round to three decimal places as needed.)

    1 answer SHARE

    Use the alternating series estimation theorem to determine how many terms should be used to estimate the sum of the entire series with an error of less than 0.001. ∑ n = 1 ∞ (−1) n + 1 1 (n+ 2 6n 3 ) 3 or more terms should be used to estimate the sum of the entire series with an error of less than 0.001 .

    1 answer SHARE

    Use the alternating series estimation theorem to determine how many terms should be used to estimate the sum of the entire series with an error of less than 0.001. ∑ n = 1 ∞ (−1) n + 1 1 (n + 5n) 3 or more terms should be used to estimate the sum of the entire series with an error of less than 0.001.

    1 answer SHARE

    Approximate the sum with an error of magnitude less than 5×10 −6 . ∑ n = 0 ∞ (−1) n 1 (3n)! The sum is approximately ◻. (Round to five decimal places as needed.)

    1 answer SHARE

    Approximate the sum with an error of magnitude less than 5×10 −7. ∑ n = 0 ∞ (−1) n 1 (5n)! The sum is approximately ◻. (Round to five decimal places as needed.)

    1 answer SHARE

    Determine whether the series ∑ n = 0 ∞ sin(18n + 1)π 2 converges or diverges. If it converges, find its sum. Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The series converges because lim n → ∞ sin(18n + 1)π 2 = 0 . The sum of the series is ◻. (Type an exact answer, using radicals as needed.) B. The series diverges because it is a geometric series with |r| ≥ 1. c. The series converges because it is a geometric series with |r| < 1. The sum of the series is ◻. (Type an exact answer, using radicals as needed.) D. The series diverges because lim n → ∞ sin(18n + 1)π 2 ≠ 0 or fails to exist. E. The series converges because lim k → ∞ ∑ n = 0 k sin(18n + 1)π 2 fails to exist.

    1 answer SHARE

    Does the series ∑ n = 1 ∞ (−1) n + 1 n 3 n 6 + 1 converge absolutely, converge conditionally, or diverge? Choose the correct answer below and, if necessary, fill in the answer box to complete your choice. A. The series converges absolutely because the limit used in the nth-Term Test is ◻. B. The series diverges because the limit used in the Ratio Test is not less than or equal to 1. C. The series converges conditionally per the Alternating Series Test and the Comparison Test with ∑ n = 1 ∞ 1 n 3. D. The series converges conditionally per the Alternating Series Test and because the limit used in the Root Test is ◻. E. The series converges absolutely per the Comparison Test with ∑ n = 1 ∞ 1 n 3. F. The series diverges because the limit used in the nth-Term Test is not zero.

    1 answer SHARE

    Does the series ∑ n = 1 ∞ (−1) n 1 n+6 converge absolutely, converge conditionally, or diverge? Choose the correct answer below and, if necessary, fill in the answer box to complete your choice. A. The series diverges because the limit used in the nth-Term Test is not zero. B. The series converges absolutely since the corresponding series of absolute values is geometric with |r| = ◻. C. The series converges conditionally per the Alternating Series Test and the Limit Comparison Test with ∑ n = 1 ∞ 1 n. D. The series converges absolutely since the corresponding series of absolute values is the p-series with p = ◻. E. The series diverges because the limit used in the Ratio Test is not less than or equal to 1. F. The series converges conditionally because the limit used in the Ratio Test is ◻.

    1 answer SHARE

    Consider the series ∑ n = 2 ∞ x n n(ln n) 3. (a) Find the series' radius and interval of convergence. (b) For what values of x does the series converge absolutely? (c) For what values of x does the series converge conditionally? (a) Find the interval of convergence. x Find the radius of convergence. R = ◻ (b) For what values of x does the series converge absolutely? x (c) For what values of x does the series converge conditionally? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The series converges conditionally at x = ◻. (Use a comma to separate answers as needed.) B. The series does not converge conditionally.

    1 answer SHARE
    • Submit Questions
    doubtrix Logo

    Doubtrix Education Help Services is one of the world’s premier online education services. The mission of our company is to provide accurate and detailed solutions.

    Quick Help
    • Ask An Expert?
    • About Us
    • Honor Code
    • Pricing & return policy
    • Assignment Solutions
    Study Help
    • Ask Your Doubt
    • Electrical Engineering
    • Math
    • Physics
    • Chemistry
    get in touch

    65, Gayatri Nagar-B, Maharani Farm, Durgapura, Jaipur-302018

    +91-6367441917
    E-Mail
    Copyright © 2021-24 Doubtrix | All Rights Reserved | Powered by GIT Infosys