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Chapter 8: Problem 40

Estimate the value of the following convergent series with an absolute error less than \(10^{-3}\). $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{3}}$$

Short Answer

Expert verified
Question: Estimate the value of the alternating convergent series with the general term $$\frac{(-1)^{k}}{(2 k+1)^3}$$ such that the absolute error is less than $10^{-3}$. Answer: The estimated value of the convergent series with an absolute error of less than $10^{-3}$ is approximately -0.12026.

Step by step solution

01

Understand Alternating Series Estimation Theorem

The Alternating Series Estimation Theorem states that for an alternating series $$\sum_{k=1}^{\infty} (-1)^{k}a_k, a_k > 0$$ if the absolute error between the partial sum and the actual sum is $$R_n = |S - S_n| \leq a_{n+1}$$ for all n, then we can use the next term in the sequence \(a_{n+1}\) as an upper bound on the error.
02

Determine required a_n+1 for given error bound.

Given that the absolute error should be less than \(10^{-3}\). We must find the minimum value of n such that: $$a_{n+1} \leq 10^{-3}$$ Substituting the general term in our series, we get: $$\frac{1}{(2(n+1) +1)^3} \leq 10^{-3}$$
03

Solve inequality for n

Now, it's time to solve the inequality for n: $$\frac{1}{(2(n+1) +1)^3} \leq 10^{-3}$$ $$\Rightarrow (2(n+1) +1)^3 \geq 1000$$ $$\Rightarrow n \geq \frac{\sqrt[3]{1000} - 3}{2} - 1$$
04

Calculate n and find the partial sum

Evaluating the expression above, we get $$n \approx 3.71$$ Since n must be an integer, we round up to the nearest integer: $$n = 4$$ Now we will find the partial sum \(S_4\) using the first 4 terms of the series: $$S_4 = \sum_{k=1}^{4} \frac{(-1)^{k}}{(2 k+1)^3}$$ $$= \frac{(-1)^1}{(2(1)+1)^3} + \frac{(-1)^2}{(2(2)+1)^3} + \frac{(-1)^3}{(2(3)+1)^3} + \frac{(-1)^4}{(2(4)+1)^3}$$ $$= -\frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + \frac{1}{9^3}$$
05

Approximate the sum

Evaluating the partial sum, we find the approximate value of the convergent series: $$S_4 \approx -0.12026$$ We can say that the absolute error in this approximation is less than \(10^{-3}\). Therefore, the estimation of the given convergent series with an absolute error of less than \(10^{-3}\) is approximately -0.12026.

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Key Concepts

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

Convergent Series
A convergent series is a series of numbers whose sequence of partial sums approaches a specific finite value, commonly referred to as the limit. This means that as you add more terms of the series, the sum gets closer and closer to a particular number. For the series given in the exercise, \( \sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2k+1)^3} \), it is an example of an alternating series. Alternating series typically flip between positive and negative terms. Convergent series are significant in calculus and analysis because they guarantee the presence of a limit. They allow us to work with infinite processes in a controlled way. The importance of discovering whether a series is convergent lies in ensuring that calculations with the series lead to meaningful and specific results, rather than diverging to infinity or getting stuck within oscillations without settling on a particular value.
Error Estimation
Error estimation in the context of a convergent series is a measure of how far an approximation of a series' sum is from its exact limit. When working with infinite series, it is often impossible or impractical to compute an infinite number of terms. Therefore, error estimation becomes crucial in determining how many terms are necessary to achieve a given level of accuracy. In the case of the provided exercise, the Alternating Series Estimation Theorem helps us estimate the maximum error. This theorem suggests that for an alternating series, the error in approximating the series using its \(n\)-th partial sum is at most the absolute value of the first omitted term, \(a_{n+1}\). Therefore, if we want the error to be less than \(10^{-3}\), we determine how large \(n\) needs to be to meet this criterion. Through this calculation, we ensure that our partial sum is a good approximation of the series' true sum and that our estimation falls within an acceptable range.
Partial Sum
A partial sum is the sum of a finite number of initial terms of a series. It provides a way to approximate an infinite series by adding together only a limited number of its terms. When given a series like \( \sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2k+1)^3} \), we can't directly sum all infinite terms. Instead, we compute partial sums, denoted as \( S_n \). In the provided step-by-step solution, we calculated the partial sum \( S_4 \), which includes the first four terms of the given series. This partial sum serves as an approximation of the series' entire sum and is calculated by evaluating each term up to the fourth and adding them up:
  • The first term is \(- \frac{1}{3^3}\)
  • The second term is \( \frac{1}{5^3}\)
  • The third term is \(- \frac{1}{7^3}\)
  • The fourth term is \( \frac{1}{9^3}\)
By summing these, we find that \( S_4 \approx -0.12026 \). This value is a close approximation of the series sum, with an error smaller than the predetermined threshold \(10^{-3}\). Thus, calculating partial sums allows us to approximate and understand infinite series through manageable steps.

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Most popular questions from this chapter

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{1}{n}=0$$

Marie takes out a \(\$ 20,000\) loan for a new car. The loan has an annual interest rate of \(6 \%\) or, equivalently, a monthly interest rate of \(0.5 \% .\) Each month, the bank adds interest to the loan balance (the interest is always \(0.5 \%\) of the current balance), and then Marie makes a \(\$ 200\) payment to reduce the loan balance. Let \(B_{n}\) be the loan balance immediately after the \(n\) th payment, where \(B_{0}=\$ 20,000\). a. Write the first five terms of the sequence \(\left\\{B_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{B_{n}\right\\}\). c. Determine how many months are needed to reduce the loan balance to zero.

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

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd .} \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N\), the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7\), and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$
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