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Chapter 8: Problem 35

Determine whether the sequence is increasing, decreasing or neither. $$a_{n}=\frac{e^{n}}{n}$$

Short Answer

Expert verified
The sequence \(a_{n}=\frac{e^{n}}{n}\) is increasing.

Step by step solution

01

Form the ratio of consecutive terms

Form the ratio of successive terms \( \frac{a_{n+1}}{a_n} \). Here, \[ \frac{a_{n+1}}{a_n} = \frac{ \frac{e^{n+1}}{n+1} }{ \frac{e^n}{n} } \] which simplifies to \[ \frac{e}{1-\frac{1}{n+1}} \]
02

Find Limit as n approaches infinity

Next, determine the limit of the ratio as \(n\) tends to infinity. From the previous step we have \[ \lim_{{n \to \infty}} \frac{e}{1-\frac{1}{n+1}} = e \]
03

Analyze the result

If the limit of the ratio of successive terms is greater than 1, the series is increasing. If the limit is less than 1, then the series is decreasing. Since our obtained limit is \(e\) which is approximately 2.71828, thus greater than 1, we can conclude that the sequence is Increasing.

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Key Concepts

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

Increasing and Decreasing Sequences
When we talk about sequences in calculus, they can exhibit different behaviors as they progress from term to term. Two such behaviors are when a sequence is termed 'increasing' or 'decreasing'.

An increasing sequence means that as you move forward from one term to the next, the value of the terms gets larger. To determine this for the given sequence, \(a_n = \frac{e^n}{n}\), we observe how the terms change as \(n\) increases. By comparing consecutive terms, \(a_{n+1}\) and \(a_n\), we gain insight into this behavior.

A decreasing sequence is the opposite, where the terms get smaller as \(n\) increases. If this pattern holds true for all terms in the sequence after a certain point, the sequence is said to be monotonically increasing or decreasing respectively. In our original exercise, we discovered that the sequence \(a_n\) is increasing since each term is larger than the preceding term as \(n\) gets larger.
Limit of a Sequence
The concept of the limit of a sequence is a fundamental part of calculus, which helps us understand the behavior of sequences as the terms approach infinity. The limit describes the value that the sequence approaches as the index \(n\) gets very large.

For the sequence in our exercise, we were tasked to find the limit of the ratio of consecutive terms as \(n\) approaches infinity, which is represented by \( \lim_{{n \to \infty}} \frac{e}{1-\frac{1}{n+1}} \). After simplifying, it was determined that the limit is \(e\), which is about 2.71828. This value tells us the trend of the sequence鈥攊n this case, since \(e > 1\), it indicates that the sequence increases without bound.
Ratio of Consecutive Terms
Another key concept in analyzing sequences is the ratio of consecutive terms, which can be used to determine whether a sequence is increasing or decreasing. This is done by dividing the term \(a_{n+1}\) by the preceding term \(a_n\).

In our example, the ratio \( \frac{a_{n+1}}{a_n} \) simplifies to \( \frac{e}{1-\frac{1}{n+1}} \) after substituting the given sequence formula. If this ratio is greater than 1 for all \(n\), the sequence is increasing, if it is less than 1, then it is decreasing. This insight is critical in understanding sequences and their long-term behavior, and it is particularly useful when the sequence's pattern is not immediately obvious.

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Most popular questions from this chapter

Use a known Taylor series to find the Taylor series about \(c=0\) for the given function and find its radius of convergence. $$f(x)=e^{-3 x}$$

$$\text { Show that } \sum_{k=2}^{\infty} \frac{1}{(\ln k)^{\ln k}} \text { and } \sum_{k=2}^{\infty} \frac{1}{(\ln k)^{k}} \text { both converge. }$$

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The energy density of electromagnetic radiation at wavelength \(\lambda\) from a black body at temperature \(T\) (degrees Kelvin) is given by Planck's law of black body radiation: \(f(\lambda)=\frac{8 \pi h c}{\lambda^{5}\left(e^{h c / 4 k T}-1\right)},\) where \(h\) is Planck's constant, \(c\) is the speed of light and \(k\) is Boltzmann's constant. To find the wavelength of peak emission, maximize \(f(\lambda)\) by minimizing \(g(\lambda)=\lambda^{5}\left(e^{h c / \lambda k T}-1\right) .\) Use a Taylor polynomial for \(e^{x}\) with \(n=7\) to expand the expression in parentheses and find the critical number of the resulting function. (Hint: Use \(\frac{h c}{k} \approx 0.014\).) Compare this to Wien's law: \(\lambda_{\max }=\frac{0.002898}{T} .\) Wien's law is accurate for small \(\lambda .\) Discuss the flaw in our use of Maclaurin series.

The power of a reflecting telescope is proportional to the surface area \(S\) of the parabolic reflector, where \(S=\frac{8 \pi}{3} c^{2}\left[\left(\frac{d^{2}}{16 c^{2}}+1\right)^{3 / 2}-1\right] .\) Here, \(d\) is the diameter of the parabolic reflector, which has depth \(k\) with \(c=\frac{d^{2}}{4 k}\) Expand the term \(\left(\frac{d^{2}}{16 c^{2}}+1\right)^{3 / 2}\) and show that if \(\frac{d^{2}}{16 c^{2}}\) is small, then \(S \approx \frac{\pi d^{2}}{4}\)
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