Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

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

Step by step solution

01

- 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.
02

- 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 \(\sum_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right) = \sum_{n=2}^{\infty} 1 - \sum_{n=2}^{\infty} \frac{1}{n^2}\).
03

- 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.
04

- 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, \(\sum_{n=2}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} - 1\).
05

- 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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the interval of convergence, including end-point tests: \(\sum_{n=1}^{\infty} \frac{x^{n} n^{2}}{5^{n}\left(n^{2}+1\right)}\)

It is useful to write series both in the form \(a_{1}+a_{2}+a_{3}+\cdots\) and in the form \(\sum_{n=1}^{\infty} a_{n} .\) Write out several terms of the following series (that is, write them in the first form). $$ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n+1} $$

Show that it is possible to stack a pile of identical books so that the top book is as far as you tike to the right of the bottom book. You may use as many books as you need, but for any desired distance it will be some finite number. Hint: Start at the top of the pile to see how to stack them. Each time place the pile already completed on top of another book so that the pile is just at the point of tipping. (In practice, of course, you can't let them overhang quite this much without having the stack topple.) Find the distance from the right-hand end of each book to the right-hand end of the one beneath it. (To find a general formula for this distance, consider the three forces acting on book \(n\), and write the equation for the torque about its right-hand end.) Show that the sum of these distances is a divergent series. (It is said that some students studying in the library tried this experiment one evening and lefe the results, to the consternation of the librarian the next morning. It is suggested that you try it instead with one or more decks of cards. Also see "Leaning Tower of The Physical Reciens," American Journal of Physics, vol. 27, no. 2, p. 12 t.)

\(\sum_{n=1}^{\infty} \frac{n^{5}}{5^{n}}\)

Use the methods of this section to find the first few terms of the Maclaurin series for each of the following functions. $$ \arcsin x=\int_{0}^{x} \frac{d t}{\sqrt{1-t^{2}}} $$

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free