/*! 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 34 Determine how many terms of the ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than \(10^{-4}\) in magnitude. Although you do not need it, the exact value of the series is given in each case. $$\frac{\pi^{3}}{32}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{3}}$$

Short Answer

Expert verified
Answer: 20 terms

Step by step solution

01

Write down the given series

The given convergent series is: $$\frac{\pi^{3}}{32}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{3}}$$ It is an alternating series, as the terms change their sign due to the factor \((-1)^k\).
02

Apply the Alternating Series Remainder Theorem

The Alternating Series Remainder Theorem states that for an alternating series, the absolute value of the remainder (the difference between the sum of the first n terms and the exact value of the series) is less than or equal to the absolute value of the \((n+1)\)-th term. In our case, the remainder can be given by: $$R_n=\left|\sum_{k=n+1}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{3}}\right|$$ We want to find \(n\) such that \(R_n < 10^{-4}\).
03

Set up the inequality

We must find the smallest integer \(n\) such that the absolute value of the \((n+1)\)-th term of the series is less than \(10^{-4}\): $$\left|\frac{(-1)^{1+n}}{(2(1+n)+1)^{3}}\right| < 10^{-4}$$
04

Simplify the inequality

Since we are only dealing with the absolute value, we don't need to consider the \((-1)^{1+n}\) factor. The inequality becomes: $$\frac{1}{(2(1+n)+1)^{3}} < 10^{-4}$$
05

Solve for n

Raise both sides of the inequality to the power of \(-1/3\) to isolate the term in the denominator: $$(2(1+n)+1) > 10^{4/3}$$ Now we can solve for \(n\): $$2(1+n) > 10^{4/3} - 1$$ $$1+n > \frac{10^{4/3} - 1}{2}$$ $$n > \frac{10^{4/3} - 1}{2} - 1$$
06

Find the smallest integer value of n

The smallest integer value of \(n\) that satisfies the inequality is: $$n = \left\lceil\frac{10^{4/3} - 1}{2} - 1\right\rceil = 20$$ So, 20 terms of the series must be summed to ensure the remainder is less than \(10^{-4}\) in magnitude.

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Key Concepts

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

Alternating Series
An alternating series is a sequence of numbers that switch between positive and negative values, typically indicated by a \textbf{multiplier of \( (-1)^n \)}. This sign-changing pattern is significant in determining the series' behavior and convergence properties. In the exercise, you encountered the alternating series \( \frac{\pi^{3}}{32} = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{3}} \), where the sign of each term alternates due to the \( (-1)^k \) term.

Understanding alternating series is crucial as it comes with particular convergence tests that might differ from series with only positive terms. This characteristic plays a fundamental role when applying the Alternating Series Remainder Theorem, as the alternating nature helps to establish bounds on the possible error – or remainder – when summing the first \( n \) terms of the series.
Alternating Series Remainder Theorem
The Alternating Series Remainder Theorem is your key tool when you need to approximate the sum of an alternating series and measure the accuracy of your approximation. It stipulates that if you sum the first \( n \) terms of a convergent alternating series, the absolute error of this partial sum (compared to the infinite sum) will be less than the absolute value of the first omitted term. Mathematically speaking, if \( R_n \) is the remainder (or error) after \( n \) terms, then \( R_n \leq |a_{n+1}| \) where \( a_{n+1} \) is the \( (n+1) \)th term.

This theorem conveniently provides a straightforward way to manage and control the error based on how many terms you are summing. As demonstrated in the exercise, by utilizing this theorem, it's possible to determine precisely how many terms of the alternating series you need to sum to achieve a desired level of accuracy.
Inequality Solving
Solving inequalities is a fundamental aspect of algebra that deals with finding the values of a variable that make the inequality true. Inequality solving often involves similar steps to solving equations, such as adding or subtracting the same terms on both sides, or multiplying or dividing both sides by the same positive number, while remembering that multiplying or dividing by a negative number reverses the inequality.

In the context of the given exercise, after applying the Alternating Series Remainder Theorem, we developed an inequality \( \frac{1}{(2(1+n)+1)^{3}} < 10^{-4} \) that needed to be solved for \( n \) to find out how many terms must be added to sum up to the required precision.

Isolate the Variable

The process included algebraic manipulations aiming to isolate \( n \) on one side of the inequality. This involved raising both sides to the power of \( -1/3 \) and arranging the terms to solve for \( n \) effectively.
Series Convergence
The concept of series convergence is one of the cornerstones of mathematical analysis. A series converges if the sum of its terms approaches a specific finite number as more and more terms are added. Conversely, if the sum grows without bound or doesn't settle on a particular value, the series diverges.

Tests for Convergence: There are multiple tests to determine whether a series converges, including the Alternating Series Test, the Ratio Test, and the Root Test, among others. The Alternating Series Test, in particular, tells us that an alternating series will converge if its terms decrease in absolute value and approach zero.

In our exercise, the fact that the series \( \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{3}} \) is given as convergent means that the sums of the series are bound to settle on the value \( \frac{\pi^{3}}{32} \). The challenge was not to demonstrate the convergence, which is a relief, but to utilize it in conjunction with the Alternating Series Remainder Theorem to find the number of terms needed for a particular precision.

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Most popular questions from this chapter

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