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Chapter 9: Problem 28

Determine whether the following series converge. $$\sum_{k=1}^{\infty}(-1)^{k} k \sin \frac{1}{k}$$

Short Answer

Expert verified
Question: Determine whether the following alternating series converges or diverges: $$\sum_{k=1}^{\infty}(-1)^{k} k \sin \frac{1}{k}$$ Answer: The given alternating series converges.

Step by step solution

01

Show that the series is alternating

To show that the given series is alternating, we must demonstrate that the series has positive and negative terms that alternate. Observe that \((-1)^k\) has different sign depending on the value of \(k\). If the value of \(k\) is odd, then \((-1)^k\) is negative, and if the value of \(k\) is even, then \((-1)^k\) is positive. Furthermore, \(\sin \frac{1}{k}\) is positive, since \(0 < \frac{1}{k} < \pi\) for all \(k\). Hence, the given series is alternating.
02

Check the limit of the absolute values of the terms

To check the limit of the absolute values of the terms, find the limit of the sequence \(a_k = |k \sin \frac{1}{k}|\), as \(k\) approaches infinity: $$\lim_{k \to \infty} |k \sin \frac{1}{k}|$$ Since \(0 < \sin x < x\) for \(0 < x < \pi\), we have that \(0 < \frac{1}{k} < \sin \frac{1}{k} < \frac{1}{k}\) for all \(k\). Then, by applying the squeeze theorem: $$\lim_{k \to \infty} |\frac{1}{k}| = 0 \le \lim_{k \to \infty} |\sin \frac{1}{k}| \le \lim_{k \to \infty} |\frac{1}{k}| = 0$$ Hence, the limit of the absolute value of the terms is 0, which satisfies one of the conditions for the AST.
03

Check if the sequence of absolute values is decreasing

To check whether the sequence of absolute values is decreasing, we will consider the derivative of the function \(f(x) = x \sin \frac{1}{x}\). If the derivative is negative for \(x > 1\), the sequence of absolute values is decreasing. Differentiating \(f(x)\) with respect to \(x\), we get: $$f'(x) = \sin(\frac{1}{x}) - \frac{1}{x} \cos(\frac{1}{x})$$ We can see that \(-\frac{\pi}{2} < \sin(\frac{1}{x}) < \frac{\pi}{2}\) and \(0 < \frac{1}{x} < 1\) for \(x > 1\). Therefore, $$-\frac{\pi}{2} - 1 < f'(x) < \frac{\pi}{2} - 1$$ As the derivative is negative for \(x > 1\), this implies that the sequence of absolute values is decreasing.
04

Apply the Alternating Series Test

Now that we have shown that the limit of the absolute values of the terms is zero, and the sequence of absolute values is decreasing, we can apply the Alternating Series Test. According to the test, the given series converges: $$\sum_{k=1}^{\infty}(-1)^{k} k \sin \frac{1}{k}$$

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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 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 remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$R_{n}=\left|S-S_{n}\right|=\left|\frac{r^{n}}{1-r}\right|$$

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{2 k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

After many nights of observation, you notice that if you oversleep one night you tend to undersleep the following night, and vice versa. This pattern of compensation is described by the relationship $$x_{n+1}=\frac{1}{2}\left(x_{n}+x_{n-1}\right), \quad \text { for } n=1,2,3, \ldots$$ where \(x_{n}\) is the number of hours of sleep you get on the \(n\) th night and \(x_{0}=7\) and \(x_{1}=6\) are the number of hours of sleep on the first two nights, respectively. a. Write out the first six terms of the sequence \(\left\\{x_{n}\right\\}\) and confirm that the terms alternately increase and decrease. b. Show that the explicit formula $$x_{n}=\frac{19}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)^{n}, \text { for } n \geq 0,$$ generates the terms of the sequence in part (a). c. What is the limit of the sequence?

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