/*! 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 24 Evaluate each geometric series o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 24

Evaluate each geometric series or state that it diverges. $$1+\frac{1}{\pi}+\frac{1}{\pi^{2}}+\frac{1}{\pi^{3}}+\cdots$$

Short Answer

Expert verified
Answer: The sum of the infinite geometric series is \(\frac{\pi}{\pi - 1}\).

Step by step solution

01

Identifying the ratio

First, we need to identify the common ratio between the terms of the geometric series. In this case, the ratio is \(\frac{1}{\pi}\). To find the ratio, we can divide the second term by the first term: $$\frac{\frac{1}{\pi}}{1} = \frac{1}{\pi}$$
02

Checking for convergence

In order for an infinite geometric series to converge, the absolute value of its ratio must be less than 1: $$|r| < 1$$ In this case, the ratio is positive: \(|\frac{1}{\pi}| = \frac{1}{\pi}\). Since \(\pi\) is greater than \(1\): $$\frac{1}{\pi} < 1$$ Therefore, the series converges.
03

Finding the sum of the series

Since the series converges, we can find its sum using the formula for the sum of an infinite geometric series: $$S = \frac{a}{1 - r}$$ where \(S\) is the sum of the series, \(a\) is the first term, and \(r\) is the common ratio. In this case, \(a = 1\) and \(r = \frac{1}{\pi}\), so we have: $$S = \frac{1}{1 - \frac{1}{\pi}}$$
04

Simplifying the sum

Now, we simplify the expression for the sum of the series: $$S = \frac{1}{\frac{\pi - 1}{\pi}} = \frac{\pi}{\pi - 1}$$ The sum of the series is: $$1+\frac{1}{\pi}+\frac{1}{\pi^{2}}+\frac{1}{\pi^{3}}+\cdots = \frac{\pi}{\pi - 1}$$

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.

Series Convergence
In mathematics, the concept of **series convergence** is fundamental when analyzing sequences and series. A series is said to converge when its terms approach a specific value as the number of terms grows indefinitely. With geometric series, specifically, the convergence depends entirely on the common ratio, denoted as \( r \). For a geometric series to converge, this ratio must satisfy the condition \( |r| < 1 \).
When this condition holds true, the terms of the series get progressively smaller, leading to a finite sum.
  • If \( |r| \geq 1 \), the series diverges, meaning it doesn鈥檛 settle to a particular finite value.
  • If \( |r| < 1 \), it converges, and we can calculate the sum using a straightforward formula.

In the example given, because the common ratio \( \frac{1}{\pi} \) is less than 1 (since \( \pi > 1 \)), the series converges, allowing further analysis to find the sum.
Infinite Series
The term **infinite series** refers to a series that continues without end, adding more and more terms indefinitely. Unlike finite series, where terms are limited, infinite series offer a continuous process of addition.
Understanding the behavior of such series is crucial in advanced mathematics and real-world applications. One of the significant challenges and goals is determining whether such a series converges or diverges.
An infinite geometric series is a particular type of infinite series where each term is derived by multiplying the previous one by a constant ratio \( r \).
  • This series can illustrate how infinite processes can effectively be "tamed" to give finite results, provided the convergence condition \( |r| < 1 \) is met.
  • In the given example, the infinite series continues indefinitely, but because \( | \frac{1}{\pi} | < 1 \), it implies it converges to a sum of \( \frac{\pi}{\pi - 1} \).
Sum of Geometric Series
The **sum of a geometric series** is a calculation that provides the total value a converging series approaches. When a geometric series meets the convergence condition \( |r| < 1 \), it is possible to determine its sum using a simple formula:
\[S = \frac{a}{1 - r}\]
Here, \( S \) represents the sum, \( a \) the first term, and \( r \) the common ratio. This formula arises from summing an infinite sequence of terms that shrink increasingly smaller.
  • The formula highlights a key feature: the sum of the series only depends on the first term and the common ratio.
  • Our example starts with \( a = 1 \) and \( r = \frac{1}{\pi} \), resulting in a sum \( S = \frac{\pi}{\pi - 1} \).
This sum shows how infinity can fold back to a finite, meaningful number through algebraic manipulation and understanding convergence.

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

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n ! \text { and } b_{n}=n^{0.7 n}, n \geq 2$$

Repeated square roots Consider the expression \(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},\) where the process continues indefinitely. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=\sqrt{1+a_{n}}\), for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\) c. Estimate the limit of the sequence. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. e. Repeat the preceding analysis for the expression \(\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}},}\) where \(p>0 .\) Make a table showing the approximate value of this expression for various values of \(p .\) Does the expression seem to have a limit for all positive values of \(p ?\)

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0},\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G.)\) a. Show that \(a_{n} > b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{7^{n}}{n^{7} 5^{n}}$$

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\int_{1}^{n} x^{-2} d x$$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

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.