Mathematics, 14.07.2021 21:10 emmapizano

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 ∑n=19∞an=∑n=19∞8n3−2n2+196+3n4 converges or dive

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Answer from: leobaut6145

Diverges

General Formulas and Concepts:

Algebra I

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

Calculus

Limits

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

Series Convergence Tests

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \sum^{\infty}_{n = 19} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}$

Step 2: Apply DCT

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

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

Step 3: Apply LCT

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

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