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

Short Answer

Expert verified
The sequence \( a_n = \frac{\ln n}{n^{1/n}} \) diverges.

Step by step solution

01

Analyze the Sequence

Examine the sequence given by \( a_n = \frac{\ln n}{n^{1/n}} \). To determine if it converges, we need to calculate the limit as \( n \to \infty \).
02

Simplify Limit Expression

Consider the expression \( n^{1/n} \). As \( n \to \infty \), we know that \( n^{1/n} \to 1 \). This is because \( \ln(n^{1/n}) = \frac{\ln n}{n} \to 0 \) as \( n \to \infty \). Thus, \( n^{1/n} \approx 1 \).
03

Calculate the Limit of the Numerator

The numerator is \( \ln n \), which approaches infinity as \( n \to \infty \). Thus, we have a form of an indeterminate ratio \( \frac{\ln n}{n^{1/n}} \approx \frac{\ln n}{1} \), but need further analysis to confirm convergence.
04

Apply L'Hopital’s Rule

To evaluate \( \lim_{n \to \infty} \frac{\ln n}{n^{1/n}} \), note initially it is not directly an indeterminate form since the denominator tends towards 1, not infinity. Therefore, check that combined expression by considering the sequence's behavior.
05

Consider Exponential Limit

Rewrite the limit: \[ \lim_{n \to \infty} \frac{\ln n}{n^{1/n}} = \lim_{n \to \infty} \frac{\ln n}{1} \approx \lim_{n \to \infty} \ln n. \] Since \( \ln n \to \infty \), the sequence diverges.

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Key Concepts

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

Limit Calculation
Understanding limit calculation is crucial to determine whether a sequence converges or diverges.The limit of a sequence \(\{a_n\}\) as \(n\) approaches infinity gives us this information. If the limit exists and is finite, the sequence converges; if not, it diverges.

For the sequence \(a_n = \frac{\ln n}{n^{1/n}}\), we are looking at how the expression behaves as \(n\) becomes very large. Recognizing growth patterns in both the numerator and denominator helps in understanding the overall tendency of the sequence.

In this exercise, the limit taught us that while the denominator simplifies to 1, the numerator grows without bound. Thus, the sequence essentially behaves like \(\ln n\), which has no upper limit, making the sequence diverge.
Indeterminate Forms
Indeterminate forms occur when a limit expression doesn't straightforwardly resolve to a distinct value. Common forms include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). When evaluating limits, such forms prevent direct computation. They signal the need for additional analysis or techniques.

In the exercise, the expression \(\frac{\ln n}{n^{1/n}}\) at first glance appears simple. As \(n^{1/n} \) approaches 1, and \(\ln n \to \infty\), it isn't immediately obvious how the sequence behaves. Recognizing when an expression might be an indeterminate form is a skill developed through practice. Here, the subtlety lies in the simplicity of the denominator approaching 1 while the numerator unfurls indefinitely, indicative of divergence.
L'Hopital's Rule
L'Hopital’s Rule is a useful tool for resolving indeterminate forms, specifically of types \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It works by differentiating the numerator and the denominator separately.Then the limit is recalculated with these derivatives.

In this problem, although L'Hopital's Rule might not directly apply due to the non-indeterminate initial look of \(\frac{\ln n}{n^{1/n}} \) turning into \(\frac{\ln n}{1} \), it's still key to understand why L'Hopital didn't step into our solution.

For more complex sequences, however, annotating potential use of L'Hopital's Rule and understanding when it's unsuitable furthers your grasp on limit behavior.
Exponential Limits
Exponential limits often involve expressions of exponential characteristics, requiring a keen eye on base manipulation and powers.Here, the term \(n^{1/n}\) plays a subtle but crucial role.

As \(n o \infty\), \(n^{1/n}\) gradually tips toward 1 - its exponential decay slower than the growth of \(\ln n\). This trait simplifies calculus but can be misleading.

Understanding why the base of an exponent shrinks at a deceiving pace equips students to accurately predict sequence behavior. Being aware of and identifying such limits helps avoid erroneous assumption of convergence.

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Most popular questions from this chapter

What is wrong here?: Multiply both sides of the alternating harmonic series $$S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+$$ $$\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\cdots$$ by 2 to get $$2 S=2-1+$$ $$\frac{2}{3}-\frac{1}{2}+\frac{2}{5}-\frac{1}{3}+\frac{2}{7}-\frac{1}{4}+\frac{2}{9}-\frac{1}{5}+\frac{2}{11}-\frac{1}{6}+\cdots$$ Collect terms with the same denominator, as the arrows indicate, to arrive at $$2 S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$$ The series on the right-hand side of this equation is the series we started with. Therefore, \(2 S=S\), and dividing by \(S\) gives \(2=1\). (Source: "Riemann's Rearrangement Theorem" by Stewart Galanor, Mathematics Teacher, Vol. 80, No. 8,1987, pp. \(675-681 .)\)

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 \(L ?\) $$ a_{1}=1, \quad a_{n+1}=a_{n}+(-2)^{n} $$

For what values of \(a,\) if any, do the series in converge? $$\sum_{n=3}^{\infty}\left(\frac{1}{n-1}-\frac{2 a}{n+1}\right)$$

According to a front-page article in the December \(15,1992,\) issue of the Wall Street Journal, Ford Motor Company used about 7\(\frac{1}{4}\) hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in \(1980 .\) The Japanese needed only about 3\(\frac{1}{2}\) hours. Ford's improvement since 1980 represents an average decrease of 6\(\%\) per year. If that rate continues, then \(n\) years from 1992 Ford will use about $$ S_{n}=7.25(0.94)^{n} $$ hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3\(\frac{1}{2}\) hours per vehicle, how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence \(\left\\{S_{n}\right\\}\) that is less than or equal to \(3.5 .\) b. Graph \(f(x)=7.25(0.94)^{x}\) and use Trace to find where the graph crosses the line \(y=3.5 .\)

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}-y=x, \quad y^{\prime}(0)=2 \text { and } y(0)=-1 $$

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