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Chapter 8: Problem 28

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

Short Answer

Expert verified
Answer: Yes, the series converges by the Alternating Series Test.

Step by step solution

01

Rewrite the series expression with absolute value

Define a new sequence with absolute value: $$a_k = |(-1)^{k} k \sin \frac{1}{k}| = k\left|\sin\frac{1}{k}\right|$$ The original series can now be written as an alternating series: $$\sum_{k=1}^{\infty}(-1)^{k}a_k$$
02

Show that \(a_k\) converges to zero as \(k\to\infty\)

Take the limit of the sequence \(a_k\) as \(k\) goes to infinity: $$\lim_{k\to\infty} a_k = \lim_{k\to\infty} k\left|\sin\frac{1}{k}\right|$$ We can use the squeeze theorem to show that this limit is zero. Notice that: $$0\leq \left|\sin\frac{1}{k}\right| \leq 1,$$ and $$\lim_{k\to\infty} k\cdot 0 = 0$$ So, using the squeeze theorem: $$\lim_{k\to\infty}k\left|\sin\frac{1}{k}\right|=0$$
03

Check if \(a_{k+1}\leq a_k\) for all \(k\)

To determine if the sequence \(a_k\) is decreasing, we need to check if \(a_{k+1}\leq a_k\) for all \(k\). Differentiate the function \(f(x)=x|\sin\frac{1}{x}|\): $$f'(x) = |\sin\frac{1}{x}| - \frac{x\cos\frac{1}{x}}{x^2}$$ We notice that \(f'(x)\leq 0\) when \(\frac{x\cos\frac{1}{x}}{x^2}\geq |\sin\frac{1}{x}|\). From the trigonometric identity, $$\sin^2 x + \cos^2 x = 1$$ we have $$|\sin x|\leq \sqrt{1- \cos^2 x}$$ Substitute \(x\) with \(\frac{1}{x}\), we get $$|\sin\frac{1}{x}|\leq \sqrt{1- \cos^{\displaystyle2} \frac{1}{x}}$$ Since \(0\leq x\leq 1\), the inequality \(\frac{x\cos\frac{1}{x}}{x^2}\geq |\sin\frac{1}{x}|\) holds, which implies that the function \(f(x)\) is decreasing.
04

Apply the Alternating Series Test

Since we have shown that \(a_k\) converges to zero as \(k\to\infty\) and \(a_k\) is decreasing, we can use the Alternating Series Test to determine the convergence of the series: The series $$\sum_{k=1}^{\infty}(-1)^{k} k \sin \frac{1}{k}$$ converges by the Alternating Series Test.

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Key Concepts

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

Alternating Series Test
An Alternating Series is one where the terms alternate in sign. The series in the exercise alternates because
  1. Every odd term is positive.
  2. Every even term is negative, as indicated by \( (-1)^k \).
To check if an alternating series converges, we use the Alternating Series Test, which requires:
  • The sequence \( a_k \) to converge to zero \, (\( \lim_{k \to \infty} a_k = 0 \)).
  • The sequence to be decreasing in absolute terms.
In the exercise, we confirmed \( a_k = k\left| \sin \frac{1}{k} \right| \) converges to zero by evaluating the limit.
The function was shown to be decreasing, validating the series' convergence.
Thus, both criteria were met, ensuring the series converges.
Squeeze Theorem
The Squeeze Theorem helps find the limit of a function when sandwiched between two other functions that have the same limit.
The exercise applied the Squeeze Theorem to determine \( \lim_{k \to \infty} k \left| \sin \frac{1}{k} \right| = 0 \).
Here is why:
  • The sine function is always between -1 and 1, meaning \( 0 \leq \left| \sin \frac{1}{k} \right| \leq 1 \).
  • Multiply everything by \( k \) to find: \( 0 \leq k \left| \sin \frac{1}{k} \right| \leq k \).
  • As \( k \) approaches infinity, \( k \cdot 0 = 0 \) and since we know \( k \to \infty \), the limit goes to 0.
Using the Squeeze Theorem, \( k \left| \sin \frac{1}{k} \right| \to 0 \), affirming the result.
Trigonometric Limits
Understanding limits involving trigonometric functions is crucial for evaluating series like the one in the exercise.
Key in this problem is noting \( \sin \frac{1}{k} \), where \( \frac{1}{k} \) becomes small as \( k \) grows large.
The property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
is essential because it shows how sine behaves for small inputs.
When \( x \) nears zero, \( \frac{\sin x}{x} \) closely approximates 1.
  • Hence, \( \left| \sin \frac{1}{k} \right| \) approaches \( \frac{1}{k} \) inversely, becoming smaller as \( k \) increases.
Recognizing this behavior is important to establish limits necessary for using the Alternating Series Test and Squeeze Theorem.

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Most popular questions from this chapter

The famous Fibonacci sequence was proposed by Leonardo Pisano, also known as Fibonacci, in about \(\mathrm{A.D.} 1200\) as a model for the growth of rabbit populations. It is given by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1 .\) Each term of the sequence is the sum of its two predecessors. a. Write out the first ten terms of the sequence. b. Is the sequence bounded? c. Estimate or determine \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}},\) the ratio of the successive terms of the sequence. Provide evidence that \(\varphi=(1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Use induction to verify the remarkable result that $$f_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n}-(-1)^{n} \varphi^{-n}\right).$$

A fallacy Explain the fallacy in the following argument. Let \(x=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots \cdot\) It follows that \(2 y=x+y\) which implies that \(x=y .\) On the other hand, $$ x-y=\underbrace{\left(1-\frac{1}{2}\right)}_{>0}+\underbrace{\left(\frac{1}{3}-\frac{1}{4}\right)}_{>0}+\underbrace{\left(\frac{1}{5}-\frac{1}{6}\right)}_{>0}+\cdots>0 $$ is a sum of positive terms, so \(x>y .\) Therefore, we have shown that \(x=y\) and \(x>y\)

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were calculating the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1,\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5,7, 11,13, \(\ldots\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) with terms that are non increasing in magnitude, converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n} .\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k .\) It can be shown that for \(n \geq 1\) \(\left|S-\left(S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right)\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|\) a. Interpret this inequality and explain why it is a better approximation to \(S\) than \(S_{n}\) b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)
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