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Chapter 11: Problem 79

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(23-84\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{(\ln n)^{200}}{n} $$

Short Answer

Expert verified
The sequence converges, and the limit is 0.

Step by step solution

01

Identify the General Term

The given sequence is \(a_n = \frac{(\ln n)^{200}}{n}\). We will analyze the behavior of \(a_n\) as \(n\) approaches infinity to determine convergence.
02

Apply L'H么pital's Rule

Since both the numerator \((\ln n)^{200}\) and the denominator \(n\) go to infinity as \(n \to \infty\), we can use L'H么pital's Rule. First, express the problem as a limit: \(\lim_{n \to \infty} \frac{(\ln n)^{200}}{n}\). Differentiate the numerator and denominator.
03

Differentiate the Numerator and Denominator

The first derivative of \((\ln n)^{200}\) with respect to \(n\) is \(\frac{200 (\ln n)^{199}}{n}\). The derivative of \(n\) is simply \(1\). Thus, the new limit to consider is \(\lim_{n \to \infty} \frac{200 (\ln n)^{199}}{n \cdot n}\).
04

Simplify the Expression

Simplifying \(\frac{200 (\ln n)^{199}}{n^2}\) as \(n \to \infty\), it becomes clear that \(n^2\) grows faster than \((\ln n)^{199}\). Apply L'H么pital's Rule repeatedly if necessary to further reduce the terms until the ratio clearly approaches zero.
05

Conclude on Convergence

Since \(\lim_{n \to \infty} \frac{200 (\ln n)^{199}}{n^2} = 0\) by repeated application of L'H么pital's Rule, the sequence \(a_n = \frac{(\ln n)^{200}}{n}\) converges to 0.

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Key Concepts

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

Convergence and Divergence
In mathematics, sequences describe an ordered list of numbers. Whether a sequence converges or diverges can help us understand its behavior as it progresses towards infinity. Convergence means that as we go further along the sequence, the numbers start approaching a specific value called the limit. Conversely, divergence suggests that the values keep getting larger or vary without settling down to a particular number.
For the sequence given as \( a_n = \frac{(\ln n)^{200}}{n} \), we are interested in checking its behavior as \( n \) increases indefinitely. If it converges, it will approach a certain number, here calculated to be 0, implying it settles close to zero despite increasing terms. If it diverges, it would mean the sequence either grows without bounds or oscillates endlessly without approaching a fixed limit.
L'H么pital's Rule
L'H么pital's Rule is a valuable tool when dealing with limits that present indeterminate forms, such as \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \). It allows us to differentiate the numerator and the denominator separately, simplifying the limit problem.
In our exercise, the sequence \( a_n = \frac{(\ln n)^{200}}{n} \) leads to an indeterminate form as \( n \to \infty \). L'H么pital's Rule comes in handy here.
We differentiate the numerator \((\ln n)^{200}\) to get \( \frac{200 (\ln n)^{199}}{n} \) and the denominator \( n \) gives us \( 1 \). This effectively reduces the complexity, and repeated applications of the rule help simplify the limit further. In doing so, we reach the conclusion that the limit approaches zero, confirming that the sequence converges.
Limit of a Sequence
The limit of a sequence is a foundational concept that helps us understand the long-term behavior of sequences. It refers to the number that the terms of a sequence get closer to as they progress indefinitely. Determining the limit can often tell us whether a sequence converges.
For our sequence \( a_n = \frac{(\ln n)^{200}}{n} \), repeated application of the derivative process, facilitated by L'H么pital's Rule, demonstrated that \( \lim_{n \to \infty} a_n = 0 \). This finding is significant, as it not only shows convergence but provides the precise target value that the sequence's terms are approaching.
Understanding the limit affords mathematicians a clear perspective of the sequence at the horizon of infinity, offering insights into its practical applications and implications.

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Most popular questions from this chapter

Euler's constant Graphs like those in Figure 11.8 suggest that as \(n\) increases there is little change in the difference between the sum $$1+\frac{1}{2}+\cdots+\frac{1}{n}$$ and the integral $$\ln n=\int_{1}^{n} \frac{1}{x} d x$$ To explore this idea, carry out the following steps. a. By taking \(f(x)=1 / x\) in the proof of Theorem 9 , show that $$\ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n$$ or $$0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1$$ Thus, the sequence $$ a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ is bounded from below and from above. b. Show that $$ \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n $$ and use this result to show that the sequence \(\left\\{a_{n}\right\\}\) in part (a) is decreasing. since a decreasing sequence that is bounded from below converges (Exercise 107 in Section 11.1\()\) , the numbers \(a_{n}\) defined in part (a) converge: $$1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma$$ The number \(\gamma,\) whose value is \(0.5772 \ldots,\) is called Euler's constant. In contrast to other special numbers like \(\pi\) and \(e,\) no other expression with a simple law of formulation has ever been found for \(\gamma .\)

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 \(L ?\) $$ a_{n}=\sqrt[n]{n} $$

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{t \rightarrow 0} \frac{1-\cos t-\left(t^{2} / 2\right)}{t^{4}} $$

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 \(L ?\) $$ a_{n}=n \sin \frac{1}{n} $$

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.) $$ \sum_{n=1}^{\infty} \operatorname{sech} n $$
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