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Using only the LCT, determine whether the series ³+²+ n=13n²+n²+1 converges or diverges b n=12. comparing it to the following geometric series:

Using only the LCT, determine whether the series ³+²+ n=13n²+n²+1 converges or diverges b n=12. comparing it to the following geometric series:

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Series Convergence with the Limit Comparison Test (LCT)** **Problem Statement** Using only the Limit Comparison Test (LCT), determine whether the following series converges or diverges: \[ \sum_{n=1}^{\infty} \frac{n^3 + 2n^2 + n}{3n^4 + n^2 + 1} \] Compare it to the following geometric series: \[ \sum_{n=1}^{\infty} \frac{1}{n} \] **Explanation of the Process** - LCT requires finding a comparable series (usually simpler) whose behavior (convergence/divergence) is known. - Typically, the most significant term in the numerator and denominator is used for comparison. In this case, we compare the given series with a known series to determine its convergence properties.
Transcribed Image Text:**Series Convergence with the Limit Comparison Test (LCT)** **Problem Statement** Using only the Limit Comparison Test (LCT), determine whether the following series converges or diverges: \[ \sum_{n=1}^{\infty} \frac{n^3 + 2n^2 + n}{3n^4 + n^2 + 1} \] Compare it to the following geometric series: \[ \sum_{n=1}^{\infty} \frac{1}{n} \] **Explanation of the Process** - LCT requires finding a comparable series (usually simpler) whose behavior (convergence/divergence) is known. - Typically, the most significant term in the numerator and denominator is used for comparison. In this case, we compare the given series with a known series to determine its convergence properties.
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