/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Consider the following sequences... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following sequences recurrence relations. Using a calculator, make a table with at least ten terms and determine a plausible limit of the sequence or state that the sequence diverges. $$a_{n}=\frac{1}{4} a_{n-1}-3 ; a_{0}=1$$

Short Answer

Expert verified
Answer: The sequence diverges.

Step by step solution

01

Understand the recurrence relation

The given sequence has the recurrence relation: $$a_{n}=\frac{1}{4} a_{n-1}-3$$ and initial condition \(a_0=1\). Start with \(a_0\) and compute first ten terms using this relation.
02

Calculate first ten terms

Using the recurrence relation, let's calculate the first ten terms of the given sequence: 1. \(a_1=\frac{1}{4} a_{0}-3=\frac{1}{4}(1)-3=-\frac{11}{4}\) 2. \(a_2=\frac{1}{4} a_{1}-3=\frac{1}{4}\left(-\frac{11}{4}\right)-3=-\frac{35}{16}\) 3. \(a_3=\frac{1}{4} a_{2}-3=\frac{1}{4}\left(-\frac{35}{16}\right)-3=-\frac{107}{64}\) 4. \(a_4=\frac{1}{4} a_{3}-3=\frac{1}{4}\left(-\frac{107}{64}\right)-3=-\frac{355}{256}\) 5. \(a_5=\frac{1}{4} a_{4}-3=\frac{1}{4}\left(-\frac{355}{256}\right)-3=-\frac{1223}{1024}\) 6. \(a_6=\frac{1}{4} a_{5}-3=\frac{1}{4}\left(-\frac{1223}{1024}\right)-3=-\frac{3877}{4096}\) 7. \(a_7=\frac{1}{4} a_{6}-3=\frac{1}{4}\left(-\frac{3877}{4096}\right)-3=-\frac{11963}{16384}\) 8. \(a_8=\frac{1}{4} a_{7}-3=\frac{1}{4}\left(-\frac{11963}{16384}\right)-3=-\frac{37039}{65536}\) 9. \(a_9=\frac{1}{4} a_{8}-3=\frac{1}{4}\left(-\frac{37039}{65536}\right)-3=-\frac{115195}{262144}\) 10. \(a_{10}=\frac{1}{4} a_{9}-3=\frac{1}{4}\left(-\frac{115195}{262144}\right)-3=-\frac{352363}{1048576}\)
03

Analyze the behavior of the sequence

Upon examining the terms, we can clearly observe that the terms of the sequence are becoming increasingly negative, with the absolute value of each term growing larger. The sequence appears to be diverging to negative infinity. Therefore, the sequence does not have a plausible limit, and we can conclude that it diverges.

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Key Concepts

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

Sequence Convergence
When studying sequences in mathematics, one of the crucial concepts is understanding when a sequence converges. Convergence refers to the behavior of a sequence as the term number gets very large, typically as it approaches infinity. A sequence converges if and only if its terms approach a specific number, known as the limit.

For a sequence to converge, the difference between the terms and the limit must become arbitrarily small as the sequence progresses. Mathematically, for any real number \( \epsilon > 0 \), there exists a positive integer \( N \) such that for all terms \( a_n \) of the sequence with \( n > N \), the inequality \( |a_n - L| < \epsilon \) is satisfied, where \(L\) denotes the limit. If such an \(L\) exists, we say that the sequence converges to \(L\).

In terms of the sequence in the given exercise, we would be looking for a number \(L\) such that as \(n\) increases, the terms \(a_n\) come arbitrarily close to \(L\). In practice, calculating the first few terms of a sequence is often a first step to observe a pattern and to hypothesize about convergence and potential limits. However, certain sequences can be deceiving, showing a trend for the initial terms but changing behavior later. Hence, a more thorough analysis is sometimes necessary.
Divergent Sequence
In contrast to convergence, a sequence is said to be divergent if it does not approach a specific limit as the term number grows. Divergence can occur in several different ways. Sometimes the terms of a sequence might grow without bound, either positively or negatively, or they might oscillate indefinitely without settling down to a single value.

An example of a divergent sequence is one where its terms keep increasing or decreasing beyond any fixed bound. In the given exercise, after calculating the first ten terms of the sequence \(a_n\), we can observe that the terms are becoming increasingly negative. The absolute values of the terms increase, suggesting that the terms are diverging to negative infinity. This behavior implies that the terms of the sequence are not approaching any real number limit, thus classifying this sequence as divergent.

Recognizing a divergent sequence early on can prevent unnecessary calculations in search for a non-existent limit. Knowing that some sequences can behave erratically without approaching a single value is key to understanding the wide array of possible behaviors of sequences.
Limit of a Sequence
The limit of a sequence is fundamentally tied to the notion of sequence convergence. When we say that a sequence has a limit, we imply that, as we progress through the sequence, its terms get infinitely close to a particular value. That value, should it exist, is called the limit of the sequence.

To say that a sequence \(a_n\) has a limit \(L\), we write \(\lim_{n \to \infty} a_n = L\). This notation encapsulates the idea that as \(n\) becomes very large, \(a_n\) approaches the value \(L\). For the sequence limit to exist, the sequence must be convergent, and every convergent sequence must have a limit—this is a two-way relationship that holds true in the realm of sequences.

However, not all sequences have limits. If a sequence keeps bouncing between different values or grows without any bound, the limit does not exist, which is often abbreviated as DNE. As seen in the problem example, where the terms grow increasingly negative with no bound, the sequence does not have a limit, reinforcing its divergent nature.

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Most popular questions from this chapter

Repeated square roots Consider the expression \(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},\) where the process continues indefinitely. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=\sqrt{1+a_{n}}\), for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\) c. Estimate the limit of the sequence. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. e. Repeat the preceding analysis for the expression \(\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}},}\) where \(p>0 .\) Make a table showing the approximate value of this expression for various values of \(p .\) Does the expression seem to have a limit for all positive values of \(p ?\)

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. Population growth 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 \(p_{0}=250\).

Determine whether the following statements are true and give an explanation or counterexample. a. \(\sum_{k=1}^{\infty}\left(\frac{\pi}{e}\right)^{-k}\) is a convergent geometric series. b. If \(a\) is a real number and \(\sum_{k=12}^{\infty} a^{k}\) converges, then \(\sum_{k=1}^{\infty} a^{k}\) converges. If the series \(\sum_{k=1}^{\infty} a^{k}\) converges and \(|a|<|b|,\) then the series \(\sum_{k=1}^{\infty} b^{k}\) converges. d. Viewed as a function of \(r,\) the series \(1+r^{2}+r^{3}+\cdots\) takes on all values in the interval \(\left(\frac{1}{2}, \infty\right)\) e. Viewed as a function of \(r,\) the series \(\sum_{k=1}^{\infty} r^{k}\) takes on all values in the interval \(\left(-\frac{1}{2}, \infty\right)\)

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

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