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Σ(-1) -1. n=1 1 4η + 1

Σ(-1) -1. n=1 1 4η + 1

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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show that the series convegres absolutely, converegs condtionally or diveregs

This image contains a mathematical series written in summation notation. The series is expressed as follows: \[ \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{1}{4n+1} \] ### Explanation: - **Summation Symbol (\(\sum\))**: This symbol denotes the sum of a sequence of terms. - **Limits of Summation**: The limits are from \(n = 1\) to \(\infty\), indicating that the summation starts at \(n = 1\) and continues indefinitely. - **Term Expression**: For each \(n\), the term being summed is \((-1)^{n-1} \cdot \frac{1}{4n+1}\). ### Breakdown of the Term: - **\((-1)^{n-1}\)**: This factor alternates the sign of each term. For \(n = 1\), it gives \((-1)^0 = 1\), for \(n = 2\), it gives \((-1)^1 = -1\), and so on. - **\(\frac{1}{4n+1}\)**: This fraction depends on the value of \(n\). As \(n\) increases, the denominator \(4n+1\) increases, making the term smaller. This series is an alternating series due to the factor \((-1)^{n-1}\), and it converges based on criteria for the convergence of alternating series. It can be used in various mathematical contexts, including analysis and series approximation.
Transcribed Image Text:This image contains a mathematical series written in summation notation. The series is expressed as follows: \[ \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{1}{4n+1} \] ### Explanation: - **Summation Symbol (\(\sum\))**: This symbol denotes the sum of a sequence of terms. - **Limits of Summation**: The limits are from \(n = 1\) to \(\infty\), indicating that the summation starts at \(n = 1\) and continues indefinitely. - **Term Expression**: For each \(n\), the term being summed is \((-1)^{n-1} \cdot \frac{1}{4n+1}\). ### Breakdown of the Term: - **\((-1)^{n-1}\)**: This factor alternates the sign of each term. For \(n = 1\), it gives \((-1)^0 = 1\), for \(n = 2\), it gives \((-1)^1 = -1\), and so on. - **\(\frac{1}{4n+1}\)**: This fraction depends on the value of \(n\). As \(n\) increases, the denominator \(4n+1\) increases, making the term smaller. This series is an alternating series due to the factor \((-1)^{n-1}\), and it converges based on criteria for the convergence of alternating series. It can be used in various mathematical contexts, including analysis and series approximation.
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