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Chapter 11: Problem 73

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(23-84\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{n^{2}}{2 n-1} \sin \frac{1}{n} $$

Short Answer

Expert verified
The sequence converges to \( \frac{1}{2} \).

Step by step solution

01

Analyze the Sine Component

The sequence terms involve the function \( \sin \frac{1}{n} \). As \( n \to \infty \), \( \frac{1}{n} \to 0 \), and thus \( \sin \frac{1}{n} \approx \frac{1}{n} \) because the sine function behaves linearly near zero. Thus, this part approximately becomes \( \frac{1}{n} \).
02

Simplify the Sequence Term

Given the approximation \( \sin \frac{1}{n} \approx \frac{1}{n} \), substitute this into the sequence term to get \( a_n \approx \frac{n^2}{2n-1} \cdot \frac{1}{n} = \frac{n}{2n-1} \).
03

Determine the Limit of the Simplified Sequence

Simplify \( a_n = \frac{n}{2n-1} \) by dividing both the numerator and the denominator by \( n \): \( a_n = \frac{1}{2 - \frac{1}{n}} \). As \( n \to \infty \), \( \frac{1}{n} \to 0 \), hence \( a_n \to \frac{1}{2} \).
04

Conclusion about Convergence

Since the limit of the simplified sequence \( \frac{n}{2n-1} \) as \( n \to \infty \) is \( \frac{1}{2} \), the original sequence \( \{ a_n \} \) converges. The limit of the sequence is \( \frac{1}{2} \).

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Key Concepts

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

Limit of a Sequence
The concept of the limit of a sequence is fundamental when discussing convergence.
In simple terms, the limit of a sequence \( \{a_n\} \) as \( n \to \infty \) is the value that the terms of the sequence approach as \( n \) becomes very large.
If a sequence has a limit, it means that the terms eventually get arbitrarily close to some fixed number. This number is called the limit.For instance, take the sequence \( a_n = \frac{n}{2n-1} \) from this exercise. We found that as \( n \to \infty \), the sequence \( a_n \) approaches \( \frac{1}{2} \).
This means that the sequence converges to \( \frac{1}{2} \).When determining the limit, always check:
  • The behavior of the terms as \( n \) grows large.
  • If the terms stabilize around a particular number.
Sequence Simplification
Simplifying a sequence is important to make it easier to find the limit.
In our exercise, we dealt with the sequence \( a_n = \frac{n^2}{2n-1} \sin \frac{1}{n} \).
The simplification process involved replacing complicated expressions with simpler equivalences to ease calculations.First, notice the function \( \sin \frac{1}{n} \), which simplifies to \( \approx \frac{1}{n} \) as \( n \to \infty \).
This simplifies our sequence term to \( \frac{n^2}{2n-1} \times \frac{1}{n} = \frac{n}{2n-1} \).Further simplification involves:
  • Dividing numerator and denominator by \( n \).
  • Resulting in \( \frac{1}{2 - \frac{1}{n}} \).
This step draws the sequence into a form where analyzing the limit is straightforward.
Sine Function Approximation
Understanding the sine function approximation is key in this exercise.
The sine function, denoted by \( \sin(x) \), behaves linearly near zero.
Specifically, when \( x \to 0 \, \sin(x) \approx x \).This characteristic was used when dealing with the sequence term \( \sin \frac{1}{n} \). As \( n \to \infty \, \frac{1}{n} \to 0 \, \sin \frac{1}{n} \approx \frac{1}{n} \).
Hence, the sine component simplifies to a form that conforms well with the sequence simplification and limit analysis.Here's the handy tip:
  • For small values of \( x \, \sin(x) \approx x \) can significantly simplify calculations.
  • Use this approximation to replace \( \sin \frac{1}{n} \) with \( \frac{1}{n} \) when \( n \) is very large.
Remember, approximations are useful when dealing with limits and make complex functions more manageable.

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Most popular questions from this chapter

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=n \sin \frac{1}{n} $$

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=(0.9999)^{n} $$

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{x \rightarrow 0} \frac{e^{x}-(1+x)}{x^{2}} $$

Outline of the proof of the Rearrangement Theorem a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n}\), and let \(s_{k}=\sum_{n=1}^{k} a_{n} .\) Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\), $$ \sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \text { and }\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2} $$ Since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\}\), there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3}\), then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1}\). Therefore, if \(n \geq N_{3}\), $$ \begin{aligned} \left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned} $$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\).

Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series鈥� convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{1}{n\left(1+\ln ^{2} n\right)} $$
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