show that the series convegres absolutely, converegs condtionally or diveregs The given image shows a mathematical expression involving an infinite sum (series). The expression is written as follows:\[ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{4n+1} \]Here's a breakdown of the components for educational purposes:- The symbol \(\sum_{n=1}^{\infty}\) denotes a summation that starts from \(n = 1\) and continues indefinitely towards infinity.- The term \((-1)^{n-1}\) indicates an alternating sign for each successive term in the series: - When \(n\) is 1, the term is \((-1)^{1-1} = 1\). - When \(n\) is 2, the term is \((-1)^{2-1} = -1\), and so on.- The fraction \(\frac{1}{4n+1}\) represents the main part of each term being summed, where \(n\) is the index of summation.In conclusion, this infinite series sums the terms of the form \((-1)^{n-1} \frac{1}{4n+1}\) starting from \(n = 1\) and continuing indefinitely.

We have applied lebnitz test to find convergence of series

Answered: • Σ(-1) -1. n=1 1 4η + 1

Math

Calculus

• Σ(-1) -1. n=1 1 4η + 1

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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Question

show that the series convegres absolutely, converegs condtionally or diveregs

The given image shows a mathematical expression involving an infinite sum (series). The expression is written as follows:

\[ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{4n+1} \]

Here's a breakdown of the components for educational purposes:

- The symbol \(\sum_{n=1}^{\infty}\) denotes a summation that starts from \(n = 1\) and continues indefinitely towards infinity.
- The term \((-1)^{n-1}\) indicates an alternating sign for each successive term in the series:
- When \(n\) is 1, the term is \((-1)^{1-1} = 1\).
- When \(n\) is 2, the term is \((-1)^{2-1} = -1\), and so on.
- The fraction \(\frac{1}{4n+1}\) represents the main part of each term being summed, where \(n\) is the index of summation.

In conclusion, this infinite series sums the terms of the form \((-1)^{n-1} \frac{1}{4n+1}\) starting from \(n = 1\) and continuing indefinitely.

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