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Chapter 10: Problem 129

Is it true that a sequence \(\left\\{a_{n}\right\\}\) of positive numbers must converge if it is bounded from above? Give reasons for your answer.

Short Answer

Expert verified
No, a sequence must not converge if it is bounded, e.g., \\( a_n = 1 + \frac{1}{n} \\) is bounded but does not converge.

Step by step solution

01

Understanding Sequence Boundedness

A sequence \( \{a_n\} \) is said to be bounded from above if there exists a number \ M \ such that \ a_n \leq M \ for all \ n \. This means all elements of the sequence are less than or equal to some fixed upper limit \ M \.
02

Convergence of a Sequence

A sequence \( \{a_n\} \) is said to converge if there exists a limit \ L \ such that for every positive number \ \epsilon \, there is a positive integer \ N \ such that for all \ n \geq N \, \ |a_n - L| < \epsilon \. This essentially means the elements of the sequence get arbitrarily close to a single value as \ n \ becomes very large.
03

Counterexample: The Bounded Harmonic Sequence

Consider the sequence \( a_n = 1 + \frac{1}{n} \). This sequence is composed of positive numbers and is bounded above by 2 (since as \ n \ increases, \ a_n \ approaches but never exceeds 2). However, \( \{a_n\} \) does not converge to any limit, showing that not all bounded sequences converge.
04

Boundedness vs Convergence

Being bounded from above does not imply convergence. A sequence can be bounded and still fail to converge if it does not approach a specific limit. The harmonic sequence \( 1 + \frac{1}{n} \) is one such example where the sequence remains bounded but does not converge as a whole.

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Key Concepts

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

Convergence
Convergence in sequences is a crucial concept in understanding the behavior of sequences and series in mathematics.
It refers to the situation where the elements of a sequence approach a single, fixed value as you move further along the sequence.

More formally, a sequence \( \{a_n\} \) converges to a limit \( L \) if for every positive number \( \epsilon \), there exists a positive integer \( N \) such that all terms of the sequence with indices \( n \geq N \) satisfy \( |a_n - L| < \epsilon \).
  • This means no matter how small an \( \epsilon \) you choose, you can find a point in the sequence beyond which all terms are closer to \( L \) than \( \epsilon \).
  • Convergence is significant in various areas of mathematics and applied fields because it ensures predictability and analytical tractability of sequences.
Understanding whether a sequence converges is often the first step in analyzing functions and their behaviors over specific intervals or domains.
Bounded Sequences
A sequence is called bounded if there is a real number that serves as a bound or limit for the elements within the sequence.
Specifically, a sequence \( \{a_n\} \) is bounded from above if there exists a number \( M \) such that \( a_n \leq M \) for every term in the sequence.

Similarly, it is bounded from below if there exists a number \( m \) such that \( a_n \geq m \) for all terms. When both these conditions are met, the sequence is simply termed as bounded.
  • Boundedness provides a useful constraint for sequences in analysis, indicating that the sequence doesn't tend toward infinity.
  • However, being bounded does not necessarily mean that a sequence will converge to a particular limit.
It's important to recognize that boundedness is often a necessary condition for convergence but not sufficient on its own. This distinction is vital when exploring the relationship between bounded sequences and their potential to converge.
Mathematical Counterexamples
Counterexamples are powerful tools in mathematics used to illustrate cases where a general rule or statement does not hold true.
They serve as concrete evidence that a certain property, such as convergence, is not guaranteed under given conditions.

For example, consider the sequence \( a_n = 1 + \frac{1}{n} \). This sequence is bounded above by 2, meaning every term is less than or equal to 2.
Nevertheless, it does not converge, because it never approaches a single, fixed limit.
  • Counterexamples like this are crucial in mathematics for validating and refining hypotheses and theorems.
  • Understanding the role of counterexamples helps students discern the boundaries of what is possible and what needs further conditions to hold true.
Using such examples aligns closely with mathematical rigor, ensuring statements are not blindly accepted without proof or exhaustive consideration of exceptional cases.

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Most popular questions from this chapter

Which of the sequences converge, and which diverge? Give reasons for your answers. \(a_{n}=1-\frac{1}{n}\)

Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2}\).

Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(\bar{x}=0\) Step 3: Calculate the ( \(n+1\) )st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=\frac{1}{\sqrt{1+x}}, \quad|x| \leq \frac{3}{4}$$

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

When \(a\) and \(b\) are real, we define \(e^{(a+i b) x}\) with the equation $$ e^{(a+i b) x}=e^{a x} \cdot e^{i b x}=e^{a x}(\cos b x+i \sin b x) $$ Differentiate the right-hand side of this equation to show that $$ \frac{d}{d x} e^{(a+i b) x}=(a+i b) e^{(a+i b) x} $$ Thus the familiar rule \((d / d x) e^{k x}=k e^{k x}\) holds for \(k\) complex as well as real.
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