Problem 10 . \(\sum_{n=2}^{\infty}\left(1-\... [FREE SOLUTION]

Chapter 1: Problem 10

. \(\sum_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)\)

Short Answer

Expert verified

The series diverges due to the \(\sum_{n=2}^{\infty} 1\) part.

Step by step solution

- Understand the series

The given series is \(\sum_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)\). This means we need to sum the terms of the form \(1-\frac{1}{n^{2}}\) starting from \(n = 2\) to infinity.

- Simplify the general term

Rewrite the general term \(1-\frac{1}{n^{2}}\). Notice that it can be split into two parts: \(1\) and \(-\frac{1}{n^2}\). The series can be written as \(\sum_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right) = \sum_{n=2}^{\infty} 1 - \sum_{n=2}^{\infty} \frac{1}{n^2}\).

- Solve the series for the first part

The series \(\sum_{n=2}^{\infty} 1\) is a divergent series because it involves summing an infinite number of terms each equal to 1. Hence, it diverges to infinity.

- Solve the series for the second part

The series \(\sum_{n=2}^{\infty} \frac{1}{n^2}\) is convergent. This is known as a p-series with \ p = 2 \ and p-series converge for \ p > 1 \. The sum of the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) is known to be \ \frac{\pi^2}{6} \. Hence, \(\sum_{n=2}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} - 1\).

- Combine the results

Combining the results we get: The infinite sum \(\sum_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right)\) involves a divergent part and a convergent part. Since the divergent part dominates, it implies that the original series is divergent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinite Series

An infinite series is a sum of infinitely many terms. These terms are generated according to a particular rule or formula. For example, in the given problem \(\sum_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)\), the terms start from \(n=2\) and continue indefinitely. Understanding whether such series converge or diverge is a crucial part of calculus and higher mathematics.

P-Series

A p-series is a specific type of infinite series of the form \(\sum_{n=1}^{\infty}\frac{1}{n^p}\). The general term of a p-series is given by \(\frac{1}{n^p}\), where \(p\) is a real number. A key property of p-series is that they converge if and only if \(p\) is greater than 1. For example, \(\sum_{n=1}^{\infty}\frac{1}{n^2}\) converges because \(p=2>1\). In the given exercise, the part \(\sum_{n=2}^{\infty}\frac{1}{n^2}\) represents a p-series with \(p=2\), which is convergent.

Divergent Series

A series is called divergent if the sum of its terms does not approach a finite limit. Instead, it continues to grow without bounds or oscillates in value. In the given exercise, \(\sum_{n=2}^{\infty}1\) is an example of a divergent series because it sums an infinite number of 1's, leading to infinity. This divergence indicates that the entire series, \(\sum_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)\), is dominated by the divergent part and hence, is divergent overall.

Convergent Series

By contrast, a series is convergent if the sum of its terms approaches a specific, finite value. For example, \(\sum_{n=2}^{\infty}\frac{1}{n^{2}}\) converges to \(\frac{\pi^2}{6}-1\). Convergent series are central in mathematics because they represent stable sums, which can be used in various applications, from physics to computer science. In this exercise, while the \(\frac{1}{n^2}\) part converges, the sum at large is driven by the divergent part, leading to an overall divergence.

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