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Chapter 9: Problem 1

Give an example of a nonincreasing sequence with a limit.

Short Answer

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Question: Provide an example of a nonincreasing sequence with a limit. Answer: The sequence \(a_n = \frac{1}{n}\) is a nonincreasing sequence with a limit of 0 as n approaches infinity.

Step by step solution

01

Identify a nonincreasing sequence

A simple example of a nonincreasing sequence is the sequence of numbers that approaches zero. A good example of this sequence is using reciprocals of natural numbers (i.e., \( \frac{1}{n} \)) . As n increases, the reciprocal value decreases. So, the sequence can be defined as \(a_n = \frac{1}{n}\).
02

Confirm that the sequence is nonincreasing

To confirm that the sequence is nonincreasing, we need to show that for every pair of consecutive terms a_n and a_(n+1), a_(n+1) is either equal to or less than a_n. Since \(a_n = \frac{1}{n}\) and \(a_{n+1} = \frac{1}{n+1}\), we need to show that: \( \frac{1}{n+1} \leq \frac{1}{n} \). As both the numerators are equal, and \(n+1 > n\), it is enough to conclude that \( \frac{1}{n+1} \leq \frac{1}{n}\), confirming that the sequence is nonincreasing.
03

Determine the limit of the sequence

Now, we need to find the limit of the sequence as n approaches infinity. The limit is represented as: \( \lim_{n \to \infty} \frac{1}{n} \). Since the denominator is increasing indefinitely while the numerator remains constant, the value of the fraction approaches zero: \( \lim_{n \to \infty} \frac{1}{n} = 0 \).
04

Conclusion

An example of a nonincreasing sequence with a limit is \(a_n = \frac{1}{n}\), which approaches a limit of 0 as n approaches infinity.

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Most popular questions from this chapter

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. When a biologist begins a study, a colony of prairie dogs has a population of \(250 .\) Regular measurements reveal that each month the prairie dog population increases by \(3 \%\) Let \(p_{n}\) be the population (rounded to whole numbers) at the end of the \(n\) th month, where the initial population is \(p_{0}=250\).

It can be proved that if a series converges absolutely, then its terms may be summed in any order without changing the value of the series. However, if a series converges conditionally, then the value of the series depends on the order of summation. For example, the (conditionally convergent) alternating harmonic series has the value $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2.$$ Show that by rearranging the terms (so the sign pattern is \(++-\) ), $$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots=\frac{3}{2} \ln 2.$$

Infinite products An infinite product \(P=a_{1} a_{2} a_{3} \ldots,\) which is denoted \(\prod_{k=1}^{\infty} a_{k}\) is the limit of the sequence of partial products \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \dots\right\\}\) a. Show that the infinite product converges (which means its sequence of partial products converges) provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges. b. Consider the infinite product $$P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \frac{24}{25} \cdots$$ Write out the first few terms of the sequence of partial products, $$P_{n}=\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)$$ (for example, \(P_{2}=\frac{3}{4}, P_{3}=\frac{2}{3}\) ). Write out enough terms to determine the value of the product, which is \(\lim _{n \rightarrow \infty} P_{n}\). c. Use the results of parts (a) and (b) to evaluate the series $$\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)$$

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(R_{n}<10^{-6}\) ). a. \(\sum_{k=0}^{\infty} 0.6^{k}\) b. \(\sum_{k=0}^{\infty} 0.15^{k}\)

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right) ; a_{0}=2, n=0,1,2, \dots$$
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