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Chapter 8: Problem 58

Find the limit of the following sequences or state that they diverge. $$\left\\{\frac{n \sin ^{3}(n \pi / 2)}{n+1}\right\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence \(\frac{n \sin ^{3}(n \pi / 2)}{n+1}\) as n approaches infinity is 0.

Step by step solution

01

Analyze the sine function

Let's first analyze the sine function part of the sequence: $$\sin ^{3}(n \pi / 2)$$ Since sine function has a period of 2蟺, \(\sin(\theta)\) repeats itself for every integer multiple of 2蟺, i.e., \(\sin(\theta) = \sin(\theta + 2k\pi)\), where k is any integer. Similarly, \(\sin ^{3}(\theta) = \sin^{3}(\theta + 2k\pi)\). Thus, \(\sin ^{3}(n \pi / 2)\) will have the same behavior as \(\sin ^{3}(\pi / 2)\). The sine function takes values between -1 and 1, and \(\sin(\pi / 2) = 1\). So, \(\sin ^{3}(n \pi /2)\) takes values between -1 and 1.
02

Observe the behavior of the entire sequence

Given the behavior of the sine function, we now analyze the entire sequence: $$\frac{n \sin ^{3}(n \pi / 2)}{n+1}$$ Since \(\sin ^{3}(n \pi /2)\) takes values between -1 and 1, we get: $$\frac{n(-1)}{n+1} \leq \frac{n \sin ^{3}(n \pi / 2)}{n+1} \leq \frac{n(1)}{n+1}$$
03

Apply the squeeze theorem

Since the numerator behaves like n, and the denominator is (n+1), the expression on the right approaches 1 as n approaches infinity. Similarly, the expression on the left approaches -1. Now, we can apply the Squeeze Theorem to find the limit as follows: $$\lim_{n \to \infty} \frac{n(-1)}{n+1} \leq \lim_{n \to \infty} \frac{n \sin ^{3}(n \pi / 2)}{n+1} \leq \lim_{n \to \infty} \frac{n(1)}{n+1}$$ $$\lim_{n \to \infty} \frac{n(1)}{n+1} = \lim_{n \to \infty} \frac{n+1-1}{n+1} = \lim_{n \to \infty} (1-\frac{1}{n+1}) = 1$$ $$\lim_{n \to \infty} \frac{n(-1)}{n+1} = \lim_{n \to \infty} (-1) = -1$$ Therefore, by the Squeeze Theorem, $$\lim_{n \to \infty} \frac{n \sin ^{3}(n \pi / 2)}{n+1}$$ exists and is equal to 0: $$\lim_{n \to \infty} \frac{n \sin ^{3}(n \pi / 2)}{n+1} = 0$$

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Key Concepts

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

Limit of a Sequence
Understanding the limit of a sequence is crucial in calculus. A sequence is a list of numbers arranged in a specific order, and the limit of a sequence is what the sequence approaches as the index (often denoted by \( n \)) becomes very large.
The concept of a limit helps us describe the behavior of sequences in the long term. It provides a way to understand whether a sequence converges (approaches a specific value) or diverges (ends up growing infinitely without settling at any particular point).
Mathematically, the limit of a sequence \( \{a_n\} \) as \( n \to \infty \) is expressed as:
  • \( \lim_{{n \to \infty}} a_n = L \), where \( L \) is a real number, if the sequence converges.
  • If no such \( L \) exists, the sequence diverges.
In the given exercise, we aim to determine whether the sequence \( \left\{ \frac{n \sin^3(n \pi / 2)}{n+1} \right\} \) converges or diverges. Through simplification and application of the Squeeze Theorem, we find that the sequence converges, with a limit of 0.
Squeeze Theorem
The Squeeze Theorem, sometimes called the Sandwich Theorem, is a useful tool for finding the limit of a sequence or function. If you have one sequence (or function) squeezed between two others, and both of those sequences or functions converge to the same limit, then the squeezed sequence will also converge to that limit.
This theorem is particularly handy when dealing with sequences or functions that oscillate or behave in complex ways. It helps deliver a clear conclusion about the limit, even if direct evaluation seems difficult.
For three functions or sequences \( f(n) \), \( g(n) \), and \( h(n) \) where:
  • \( f(n) \leq g(n) \leq h(n) \) for all sufficiently large \( n \)
  • \( \lim_{n \to \infty} f(n) = \lim_{n \to \infty} h(n) = L \)
Then, it follows that:
  • \( \lim_{n \to \infty} g(n) = L \)
In the provided solution, \( \frac{n(-1)}{n+1} \leq \frac{n \sin^3(n \pi / 2)}{n+1} \leq \frac{n(1)}{n+1} \). By applying the Squeeze Theorem, it is concluded that the limit of the target sequence is 0.
Behavior of Trigonometric Functions
Trigonometric functions, such as sine and cosine, are periodic. This means they repeat their values in regular intervals. The sine function, \( \sin(\theta) \), has a period of \( 2\pi \). This implies that the function values return to their original state every \( 2\pi \) units.
When dealing with sequences involving trigonometric functions, understanding periodicity is essential. It simplifies the complexity associated with continuous oscillation, allowing us to make predictions about a function's behavior at large values of \( n \).
In the exercise, \( \sin^3(n \pi / 2) \) is critical. Since \( \sin(n \pi / 2) \) only takes values of \( \pm1 \) (due to the sine function's periodic behavior), \( \sin^3(n \pi / 2) \) behaves similarly. Consequently, \( \sin^3(n \pi / 2) \) will also be between -1 and 1.
Recognizing this boundedness helps in setting up the squeeze limits effectively, assisting in the application of the Squeeze Theorem to the given sequence.

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Most popular questions from this chapter

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{4^{n}+5 n !}{n !+2^{n}}$$

Given any infinite series \(\sum a_{k}\) let \(N(r)\) be the number of terms of the series that must be summed to guarantee that the remainder is less than \(10^{-r}\) in magnitude, where \(r\) is a positive integer. a. Graph the function \(N(r)\) for the three alternating \(p\) -series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{p}},\) for \(p=1,2,\) and \(3 .\) Compare the three graphs and discuss what they mean about the rates of convergence of the three series. b. Carry out the procedure of part (a) for the series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k !}\) and compare the rates of convergence of all four series.

The Greek philosopher Zeno of Elea (who lived about 450 B.c.) invented many paradoxes, the most famous of which tells of a race between the swift warrior Achilles and a tortoise. Zeno argued The slower when running will never be overtaken by the quicker: for that which is pursuing must first reach the point from which that which is fleeing started, so that the slower must necessarily always be some distance ahead. In other words, by giving the tortoise a head start, Achilles will never overtake the tortoise because every time Achilles reaches the point where the tortoise was, the tortoise has moved ahead. Resolve this paradox by assuming that Achilles gives the tortoise a 1 -mi head start and runs \(5 \mathrm{mi} / \mathrm{hr}\) to the tortoise's \(1 \mathrm{mi} / \mathrm{hr}\). How far does Achilles run before he overtakes the tortoise, and how long does it take?

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}} .\) When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots$$ Use the estimation techniques described in the text to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

Consider the alternating series $$ \sum_{k=1}^{\infty}(-1)^{k+1} a_{k}, \text { where } a_{k}=\left\\{\begin{array}{cl} \frac{4}{k+1}, & \text { if } k \text { is odd } \\ \frac{2}{k}, & \text { if } k \text { is even } \end{array}\right. $$ a. Write out the first ten terms of the series, group them in pairs, and show that the even partial sums of the series form the (divergent) harmonic series. b. Show that \(\lim _{k \rightarrow \infty} a_{k}=0\) c. Explain why the series diverges even though the terms of the series approach zero.
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