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Use the limit comparison test to determine whether ∑n=19∞an=∑n=19∞8n3−2n2+196+3n4 converges or diverges.

(a) Choose a series ∑n=19∞bn with terms of the form bn=1np and apply the limit comparison test. Write your answer as a fully simplified fraction. For n≥19,

limn→∞anbn=limn→∞


(b) Evaluate the limit in the previous part. Enter ∞ as infinity and −∞ as -infinity. If the limit does not exist, enter DNE.
limn→∞anbn =


(c) By the limit comparison test, does the series converge, diverge, or is the test inconclusive?

Use the limit comparison test to determine whether n19ann198n32n21963n4 converges or diverges a Choose a series n19bn with terms of the form bn1np and apply the class=

Respuesta :

Space

Answer:

Diverges

General Formulas and Concepts:

Algebra I

  • Exponential Rule [Dividing]:                                                                         [tex]\displaystyle \frac{b^m}{b^n} = b^{m - n}[/tex]

Calculus

Limits

  • Limit Rule [Variable Direct Substitution]:                                                     [tex]\displaystyle \lim_{x \to c} x = c[/tex]

Series Convergence Tests

  • P-Series:                                                                                                         [tex]\displaystyle \sum^{\infty}_{n = 1} \frac{1}{n^p}[/tex]
  • Direct Comparison Test (DCT)
  • Limit Comparison Test (LCT):                                                                       [tex]\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n}[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \sum^{\infty}_{n = 19} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}[/tex]

Step 2: Apply DCT

  1. Define Comparison:                                                                                     [tex]\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{n^3}{n^4}[/tex]
  2. [Comparison Sum] Simplify:                                                                         [tex]\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{1}{n}[/tex]
  3. [Comparison Sum] Determine convergence:                                             [tex]\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{1}{n} = \infty , \ \text{div by P-Series}[/tex]
  4. Set up inequality comparison:                                                                     [tex]\displaystyle\frac{8n^3 - 2n^2 + 19}{6 + 3n^4} \geq \frac{1}{n}[/tex]
  5. [Inequality Comparison] Rewrite:                                                                 [tex]\displaystyle n(8n^3 - 2n^2 + 19) \geq 6 + 3n^4[/tex]
  6. [Inequality Comparison] Simplify:                                                                 [tex]\displaystyle 8n^4 - 2n^3 + 19n \geq 6 + 3n^4 \ \checkmark \text{true}[/tex]

∴ the sum  [tex]\displaystyle \sum^{\infty}_{n = 19} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}[/tex]  is divergent by DCT.

Step 3: Apply LCT

  1. Define:                                                                                                           [tex]\displaystyle a_n = \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}, \ b_n = \frac{1}{n}[/tex]
  2. Substitute in variables [LCT]:                                                                       [tex]\displaystyle \lim_{n \to \infty} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4} \cdot n[/tex]
  3. Simplify:                                                                                                         [tex]\displaystyle \lim_{n \to \infty} \frac{8n^4 - 2n^3 + 19n}{6 + 3n^4}[/tex]
  4. [Limit] Evaluate [Coefficient Power Rule]:                                                   [tex]\displaystyle \lim_{n \to \infty} \frac{8n^4 - 2n^3 + 19n}{6 + 3n^4} = \frac{8}{3}[/tex]

∴ Because  [tex]\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} \neq 0[/tex]  and the sum  [tex]\displaystyle \sum^{\infty}_{n = 19} a_n[/tex]  diverges by DCT,  [tex]\displaystyle \sum^{\infty}_{n = 19} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}[/tex]   also diverges by LCT.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Convergence Tests (BC Only)

Book: College Calculus 10e

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