/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Determine whether the series is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 1

Determine whether the series is a \(p\)-series. $$\sum_{n=1}^{\infty} \frac{1}{n^{2}}$$

Short Answer

Expert verified
Yes, the given series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) is a p-series with \(p=2\).

Step by step solution

01

Identification

The given series is \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\). It has the format of \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\) with \(p = 2\).
02

Compare with p-series

Write it in the form of \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\) where p would be the exponent of n at the denominator. In this case, \(p=2\).
03

Determine if it is a p-series

A p-series is a series of the form \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\). Therefore, the given series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) is indeed a p-series, as it has exactly this form with \(p=2\).

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 91影视!

Key Concepts

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

Infinite Series
An infinite series is essentially a sum of an infinite sequence of numbers. For example, the series

\[\begin{equation}\sum_{n=1}^{\infty} a_n\end{equation}\]

involves adding up the terms a1, a2, a3, ... consecutively. In the case of the series from the exercise,

\[\begin{equation}\sum_{n=1}^{\infty} \frac{1}{n^{2}}\end{equation}\]

we see that it's an infinite series where the term an equals \frac{1}{n^{2}}. This series adds up the reciprocals of the squares of all positive integers.
Convergence of Series
The convergence of a series is a key concept in understanding whether the sum of an infinite series reaches a finite value, or whether it grows without bound. In mathematical terms, we say a series

\[\begin{equation}\sum_{n=1}^{\infty} a_n\end{equation}\]

converges to a number L if the sequence of partial sums approaches L as more terms are added. In context with the p-series given in the exercise,

\[\begin{equation}\sum_{n=1}^{\infty} \frac{1}{n^{2}}\end{equation}\]

converges, meaning that as we infinitely add the terms \frac{1}{1^2}, \frac{1}{2^2}, \frac{1}{3^2}, and so on, the series approaches a specific finite value. For a p-series, when p > 1, the series always converges.
Series Comparison Test
The series comparison test is a method used to determine the convergence of a series by comparing it to a known benchmark series. We consider the terms of two series, say an and bn, and if for all n,

\[\begin{equation}0 \leq a_n \leq b_n\end{equation}\]

and if the sum of bn is known to converge, then the sum of an also converges. Similarly, if the sum of bn diverges, and an is always greater than bn, then the sum of an also diverges. For the exercise at hand, the comparison test is not immediately necessary since we recognize it's a convergent p-series with p=2. However, if the value of p were less clear, the comparison test could provide a useful tool to establish convergence or 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 sum of the convergent series. $$ \sum_{n=0}^{\infty}\left[(0.4)^{n}-(0.8)^{n}\right] $$

Verify that the infinite series diverges. $$ \sum_{n=1}^{\infty} \frac{n}{\sqrt{n^{2}+1}}=\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{5}}+\frac{3}{\sqrt{10}}+\frac{4}{\sqrt{17}}+\cdots $$

Verify that the infinite series diverges. $$ \sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}=1+\frac{4}{3}+\frac{16}{9}+\frac{64}{27}+\cdots $$

The repeating decimal is expressed as a geometric series. Find the sum of the geometric series and write the decimal as the ratio of two integers. $$ 0 . \overline{4}=0.4+0.04+0.004+0.0004+\cdots $$

The repeating decimal is expressed as a geometric series. Find the sum of the geometric series and write the decimal as the ratio of two integers. $$ 0 . \overline{21}=0.21+0.0021+0.000021+\cdots $$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation
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.