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Chapter 8: Problem 80

Two sine series Determine whether the following series converge. a. \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) b. \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\)

Short Answer

Expert verified
Question: Determine the convergence of the following series: a. \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) b. \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) Answer: - The series \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) is not absolutely convergent. - The series \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) is conditionally convergent.

Step by step solution

01

Use the absolute convergence

Check if the series converges absolutely by taking the absolute value of the terms: $$\sum_{k=1}^{\infty} |\sin \frac{1}{k}|$$
02

Compare with the harmonic series

We know that \(0 \le |\sin x| \le |x|\) for all x in the interval \([0,\pi/2]\). Since \(\frac{1}{k}\) decreases as \(k\) increases, and the series converges in the positive direction, we can compare the given series to the absolute value of the harmonic series: $$0 \le |\sin \frac{1}{k}| \le \frac{1}{k}$$ Notice that the harmonic series \(\sum_{k=1}^{\infty} \frac{1}{k}\) is a well-known divergent series.
03

Determine the convergence of series a

The comparison test tells us that if the larger series (\(\sum_{k=1}^{\infty} \frac{1}{k}\)) is divergent, then the smaller series in question (\(\sum_{k=1}^{\infty} |\sin \frac{1}{k}|\)) must also be divergent. Since the harmonic series diverges, the series \(\sum_{k=1}^{\infty} |\sin \frac{1}{k}|\) also diverges. Therefore, the original series \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) is not absolutely convergent. ##Series b: \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\)##
04

Apply Dirichlet's Test

We want to see if the series converges conditionally. To apply Dirichlet's Test, we must rewrite the series as a product of two sequences: $$\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k} = \sum_{k=1}^{\infty} a_k b_k$$ where \(a_k = \frac{1}{k}\) and \(b_k = \sin \frac{1}{k}\).
05

Check for conditions that satisfy Dirichlet's Test

Dirichlet's Test states that if the following two conditions are met, then the series converges: 1. The sequence \(a_k\) is a monotonically decreasing sequence of non-negative terms that converges to 0. 2. The partial sums of the sequence \(b_k\) are bounded. Since \(\frac{1}{k}\) is monotonically decreasing and converges to 0 as \(k\) goes to infinity, condition 1 is satisfied.
06

Study the partial sums of \(b_k\)

Calculate the partial sums of the sequence \(b_k\): $$S_n = \sum_{k=1}^{n} \sin \frac{1}{k}$$ Since \(-1 \le \sin x \le 1\) for all \(x\), we also have \(-1 \le \sin \frac{1}{k} \le 1\). Therefore, the sequence \(b_k = \sin \frac{1}{k}\) is bounded. Hence, the partial sums of the sequence are also bounded, satisfying condition 2.
07

Determine the convergence of series b

Because both conditions of Dirichlet's Test are met, we can conclude that the series \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) is conditionally convergent. To summarize: - The series \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) is not absolutely convergent. - The series \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) is conditionally convergent.

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Most popular questions from this chapter

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b>1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n !>b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

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 \(\lim _{n \rightarrow \infty} x_{n}=\ln 2\).

Imagine that the government of a small community decides to give a total of \(\$ W\), distributed equally, to all its citizens. Suppose that each month each citizen saves a fraction \(p\) of his or her new wealth and spends the remaining \(1-p\) in the community. Assume no money leaves or enters the community, and all the spent money is redistributed throughout the community. a. If this cycle of saving and spending continues for many months, how much money is ultimately spent? Specifically, by what factor is the initial investment of \(\$ W\) increased (in terms of \(p\) )? Economists refer to this increase in the investment as the multiplier effect. b. Evaluate the limits \(p \rightarrow 0\) and \(p \rightarrow 1,\) and interpret their meanings.

In Section \(8.3,\) we established that the geometric series \(\sum r^{k}\) converges provided \(|r| < 1\). Notice that if \(-1 < r<0,\) the geometric series is also an alternating series. Use the Alternating Series Test to show that for \(-1 < r <0\), the series \(\sum r^{k}\) converges.

Determine whether the following statements are true and give an explanation or counterexample. a. \(\sum_{k=1}^{\infty}\left(\frac{\pi}{e}\right)^{-k}\) is a convergent geometric series. b. If \(a\) is a real number and \(\sum_{k=12}^{\infty} a^{k}\) converges, then \(\sum_{k=1}^{\infty} a^{k}\) converges. If the series \(\sum_{k=1}^{\infty} a^{k}\) converges and \(|a|<|b|,\) then the series \(\sum_{k=1}^{\infty} b^{k}\) converges. d. Viewed as a function of \(r,\) the series \(1+r^{2}+r^{3}+\cdots\) takes on all values in the interval \(\left(\frac{1}{2}, \infty\right)\) e. Viewed as a function of \(r,\) the series \(\sum_{k=1}^{\infty} r^{k}\) takes on all values in the interval \(\left(-\frac{1}{2}, \infty\right)\)
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