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Chapter 9: Problem 39

Estimate the value of the following convergent series with an absolute error less than \(10^{-3}\). $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{5}}$$

Short Answer

Expert verified
Series: \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{5}}\) Answer: The approximate sum of the series is 0.999091697 with an absolute error of less than \(10^{-3}\).

Step by step solution

01

Identify series type

The given series is an alternating series, which is represented as: $$ \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{5}}$$
02

Apply the Alternating Series test

An alternating series converges when the sequence it is based on is decreasing and has a limit of 0. We can observe that the absolute value of the given series is decreasing and its limit is 0. So, the series converges.
03

Estimate the sum with an absolute error of \(10^{-3}\)

To achieve an absolute error of less than \(10^{-3}\), we must find the smallest n such that the absolute value of the (n+1)-th term is less than \(10^{-3}\). Let $$a_n = \frac{1}{n^5}$$ We need to find the smallest n for which: $$|a_{n+1}| < 10^{-3}$$
04

Solve the inequality for n

Let's solve the inequality for n. $$|a_{n+1}| = \frac{1}{(n+1)^5} < 10^{-3}$$ Now raise each side to the power of 1/5: $$(n+1)^5 > 10^3$$ $$(n+1) > 10$$
05

Evaluate the sum up to n terms

Given that \((n+1) > 10\), we can take n = 9 as the lowest value. So, the sum of the series up to n terms is approximately: $$S \approx \sum_{k=1}^{9} \frac{(-1)^{k}}{k^{5}}$$ Now, compute the sum: $$S \approx \frac{1}{1^5} - \frac{1}{2^5} + \frac{1}{3^5} - \frac{1}{4^5} + \frac{1}{5^5} - \frac{1}{6^5} + \frac{1}{7^5} - \frac{1}{8^5} + \frac{1}{9^5}$$ $$S \approx 0.999091697$$
06

Conclusion

The evaluation of the convergent series \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{5}}\) is approximately 0.999091697 with an absolute error of less than \(10^{-3}\).

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Most popular questions from this chapter

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}.\) (Assume the result of Exercise 63.)

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln k}$$

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(R_{n}<10^{-6}\) ). a. \(\sum_{k=0}^{\infty} 0.6^{k}\) b. \(\sum_{k=0}^{\infty} 0.15^{k}\)

Consider the sequence \(\left\\{x_{n}\right\\}\) defined for \(n=1,2,3, \ldots\) by $$x_{n}=\sum_{k=n+1}^{2 n} \frac{1}{k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2 n}$$ a. Write out the terms \(x_{1}, x_{2}, x_{3}\) b. Show that \(\frac{1}{2} \leq x_{n}<1,\) for \(n=1,2,3, \ldots\) c. Show that \(x_{n}\) is the right Riemann sum for \(\int_{1}^{2} \frac{d x}{x}\) using \(n\) subintervals. d. Conclude that \(\lim _{n \rightarrow \infty} x_{n}=\ln 2\)

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