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In Exercises \(97-100,\) determine if the sequence is nondecreasing and if it is bounded from above. $$ a_{n}=2-\frac{2}{n}-\frac{1}{2^{n}} $$

Short Answer

Expert verified
The sequence is not nondecreasing but is bounded above by 2.

Step by step solution

01

Define a Nondecreasing Sequence

To determine if the sequence \(a_n\) is nondecreasing, we need to verify if \(a_{n+1} \geq a_n\) for all \(n\). This implies that the sequence does not decrease as \(n\) increases.
02

Compare Consecutive Terms

Calculate the difference of consecutive terms: \[ a_{n+1} - a_n = \left(2-\frac{2}{n+1}-\frac{1}{2^{n+1}}\right) - \left(2-\frac{2}{n}-\frac{1}{2^n}\right) \]Simplify the expression:\[ = -\frac{2}{n+1} + \frac{2}{n} + \frac{1}{2^n} - \frac{1}{2^{n+1}} \]Further simplify using the fact that \(\frac{1}{2^{n+1}} = \frac{1}{2\cdot2^n}\):\[ = \frac{2n - 2(n+1)}{n(n+1)} + \frac{1}{2^n}(1 - \frac{1}{2}) \]\[ = \frac{-2}{n(n+1)} + \frac{1}{2^{n+1}} \]
03

Analyze the Difference

We see that:\[ \frac{-2}{n(n+1)} + \frac{1}{2^{n+1}} \leq 0 \]for sufficiently large \(n\), since \(\frac{1}{2^{n+1}}\) becomes very small and the negative term \(\frac{-2}{n(n+1)}\) dominates. Therefore, for large values of \(n\), the sequence may not be nondecreasing.
04

Determine if the Sequence is Bounded from Above

To determine if \(a_n\) is bounded above, investigate its limit as \(n\) approaches infinity:\[ \lim_{n \to \infty} \left(2-\frac{2}{n}-\frac{1}{2^n}\right) = 2 - 0 - 0 = 2 \]Thus, the sequence approaches 2, which acts as an upper bound. Therefore, \(a_n\) is bounded from above by 2.

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Key Concepts

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

Understanding Nondecreasing Sequences
A nondecreasing sequence is one where each term is greater than or at the very least, equal to the previous term. Essentially, as you move further in the sequence, the terms do not become smaller. Mathematically, this means that for a sequence \(a_n\), for every positive integer \(n\), the inequality \(a_{n+1} \geq a_n\) holds true.

To check if a sequence is nondecreasing, you compare each term to the next. If you find even one instance where \(a_{n+1} < a_n\), then the sequence is not nondecreasing. It is important to note that a sequence can still be nondecreasing even if some terms are equal because the key is the absence of decrease.

In the given exercise, the sequence \(a_n = 2-\frac{2}{n}-\frac{1}{2^n}\) was analyzed to determine if it is nondecreasing. Upon simplifying and examining the difference between consecutive terms, it was determined that for large values of \(n\), the sequence may cease to be nondecreasing due to the negative component in the terms' difference.
Bounding Sequences: Bounded from Above
When discussing bounded sequences, a sequence is considered to be bounded from above if there exists a number that no term in the sequence exceeds. Think of it as having a ceiling; the sequence can rise towards it but never surpass it.

To determine if a sequence is bounded from above, one needs to calculate the limit as \(n\) approaches infinity. If there is a particular value towards which the sequence is converging and which acts as a boundary, that value is the upper bound. In this exercise, the sequence \(a_n = 2 - \frac{2}{n} - \frac{1}{2^n}\) was evaluated and found to approach the value of 2 as \(n\) increases indefinitely.

Hence, 2 can be considered an upper bound because as \(n\) becomes very large, the terms \(\frac{2}{n}\) and \(\frac{1}{2^n}\) diminish towards zero, solidifying that the sequence won't exceed its limit, which is 2.
Diving into the Limits of Sequences
The concept of limits in sequences deals with identifying a point that the terms of a sequence approach as the sequence progresses. It is crucial in understanding the long-term behavior of the sequence.

To find the limit of a sequence, examine the expression for the terms as \(n\) becomes arbitrarily large. If the terms appear to hone in on a specific value, that is the limit. This provides insight into whether the sequence becomes steady.
  • A sequence has a limit \(L\) if for every tiny distance away from \(L\), no matter how small, there is a point in the sequence past which all terms are that close to \(L\).
  • Equivalent mathematical expression: \(\lim_{n \to \infty} a_n = L\).
In the exercise, the limit as \(n\) approaches infinity for \(a_n = 2 - \frac{2}{n} - \frac{1}{2^n}\) was calculated to be 2. This means as \(n\) gets larger, the terms of the sequence get closer and closer to 2, providing a stable understanding of its behavior.

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Most popular questions from this chapter

In Exercises \(121-124\) , experiment with a calculator to find a value of \(N\) that will make the inequality hold for all \(n>N\) . Assuming that the inequality is the one from the formal definition of the limit of a sequence, what sequence is being considered in each case and what is its limit? $$ 2^{n} / n !<10^{-7} $$

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{y \rightarrow 0} \frac{\tan ^{-1} y-\sin y}{y^{3} \cos y} $$

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 \(y=3.5 .\)

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

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

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