/*! 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 89 Consider the geometric series \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 89

Consider the geometric series \(S=\sum_{k=0}^{\infty} r^{k}\) which has the value \(1 /(1-r)\) provided \(|r|<1\). Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The magnitude of the remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$ R_{n}=S-S_{n}=\frac{r^{n}}{1-r} $$

Short Answer

Expert verified
Question: Show that the remainder, $R_n$, in the sum of the infinite geometric series and the sum of its first n terms can be expressed as $R_n = \frac{r^{n}}{1-r}$. Answer: We have derived the expression for the remainder, $R_n$, as $R_n = \frac{r^{n}}{1-r}$.

Step by step solution

01

Understand the given expressions and relations

Recall that the sum of an infinite geometric series is given by \(S = \frac{1}{1-r}\), when the absolute value of common ratio \(r\) is less than 1. Also, the sum of the first \(n\) terms of the geometric series, denoted by \(S_n\), is given by the formula \(S_n = \frac{1-r^n}{1-r}\). According to the exercise, we need to find the error between these values, which is the remainder \(R_n\), so \(R_n = S - S_n\).
02

Express the difference between \(S\) and \(S_n\)

We are given that \(S = \frac{1}{1-r}\) and \(S_n = \frac{1-r^n}{1-r}\). To find the difference between them, we'll subtract \(S_n\) from \(S\). So, we have: $$ R_{n} = S - S_{n} = \frac{1}{1-r} - \frac{1-r^n}{1-r} $$
03

Simplify the expression

The next step is to simplify the expression for \(R_n\). To do this, we'll find a common denominator. Notice that the denominator of \(S\) is \((1-r)\), thus we'll multiply the numerator and denominator of the first term by (1-r) so that both fractions have the same denominator: $$ R_{n} = \frac{(1-r)(1)}{1-r} - \frac{1-r^n}{1-r} $$ Now that both fractions have the same denominator, we can combine them: $$ R_n = \frac{1 - r - (1 - r^n)}{1-r} $$
04

Simplify the numerator

Notice that the "1" terms in the numerator cancel each other. Therefore, we have: $$ R_n = \frac{-r + r^n}{1-r} $$
05

Factor out the r term

We can now factor out the r term in the numerator: $$ R_n = \frac{r(r^{n-1} - 1)}{1-r} $$
06

Observe the signs

Notice that the signs within the parenthesis on the numerator are opposite from our target expression. We can fix this by multiplying the top and the bottom of the fraction by -1: $$ R_n = \frac{-r(-r^{n-1} + 1)}{-1(1-r)} $$
07

Simplify the expression

Now, we can simplify the expression to obtain the desired formula for the remainder: $$ R_n = \frac{r^{n}}{1-r} $$ Thus, we have derived the given formula for the remainder \(R_n\).

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
Infinite series are sums of infinitely many terms. They occur in a variety of math problems. Geometric series are a type of infinite series, defined by having each term as a constant ratio of the previous term. For instance, the series \( \sum_{k=0}^{\infty} r^{k} \) represents an infinite geometric series where \( r \) is the common ratio. A critical requirement is that \( |r| < 1 \) to ensure the series converges to a finite value. In layman's terms, this means the series will not explode to infinity but will settle on some number. In this infinite geometric series, the sum of all terms is calculated by the formula \( S = \frac{1}{1-r} \), reflecting the limiting value that the series approaches as more terms are added.
Sum of Series
Considering finite sums first helps understand infinite ones. The sum of the first \( n \) terms of a geometric series, known as a partial sum, is represented by \( S_n = \frac{1-r^n}{1-r} \). Why is this important? Because calculating an infinite sum directly is impractical. We approximate by summing a large number of terms.
  • \( S_n \) gives us an approximate value of the total series by just calculating those first \( n \) terms.
  • It captures how much of the infinite series we have accounted for.
As \( n \) increases, \( S_n \) becomes a better approximation of \( S \), inching closer and closer to the full sum described earlier.
Remainder Calculation
The remainder \( R_n \) tells us how much of the series is left out when approximating using \( S_n \). It's the difference between the infinite series \( S \) and the partial sum \( S_n \). This can be calculated using:\[ R_n = S - S_n = \frac{r^n}{1-r} \] Here's why it matters:
  • Understanding \( R_n \) helps in knowing the error made when approximating an infinite series with a finite one.
  • It quantifies how close \( S_n \) is to \( S \). Less remainder means a more accurate approximation.
This calculation showcases how each additional term in the series improves the estimation, getting us ever more precise in representing the sum of the infinite series.

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

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)},$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). a. Explain why \(\left\\{F_{n}\right\\}\) is a decreasing sequence. b. Plot \(\left\\{F_{n}\right\\},\) for \(n=1,2, \ldots, 20\). c. Based on your experiments, make a conjecture about \(\lim _{n \rightarrow \infty} F_{n}\).

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} 9(0.1)^{k}$$

$$\text {Evaluate each series or state that it diverges.}$$ $$\sum_{k=1}^{\infty}\left(\sin ^{-1}(1 / k)-\sin ^{-1}(1 /(k+1))\right)$$

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 \(1 / \mathrm{AGM}(1, \sqrt{2})\).

The Greek philosopher Zeno of Elea (who lived about 450 B.c.) invented many paradoxes, the most famous of which tells of a race between the swift warrior Achilles and a tortoise. Zeno argued The slower when running will never be overtaken by the quicker: for that which is pursuing must first reach the point from which that which is fleeing started, so that the slower must necessarily always be some distance ahead. In other words, by giving the tortoise a head start, Achilles will never overtake the tortoise because every time Achilles reaches the point where the tortoise was, the tortoise has moved ahead. Resolve this paradox by assuming that Achilles gives the tortoise a 1 -mi head start and runs \(5 \mathrm{mi} / \mathrm{hr}\) to the tortoise's \(1 \mathrm{mi} / \mathrm{hr}\). How far does Achilles run before he overtakes the tortoise, and how long does it take?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.