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Chapter 2: Problem 13

If a monotonic sequence \(\left(a_{n}\right)\) has a subsequence \(\left(a_{n_{k}}\right)\) such that \(a_{n_{k}} \rightarrow a\) where \(a \in \mathbb{R}\) or \(a=\infty\) or \(a=-\infty\), then show that \(a_{n} \rightarrow a\).

Short Answer

Expert verified
Since the given sequence is monotonic, it is either increasing or decreasing. In both cases, we can use the convergence of the subsequence \((a_{n_k})\) to the limit \(a\) and the monotonic property of the sequence to show that the whole sequence \((a_n)\) also converges to the same limit \(a\). For an increasing sequence, we have \(|a_n - a| = a - a_n \leq a - a_{n_K} < \epsilon\) for all \(n \geq n_K\), while for a decreasing sequence, we have \(|a_n - a| = a_n - a \leq a_{n_K} - a < \epsilon\) for all \(n \geq n_K\). Therefore, in both cases, it is proven that \(a_n \rightarrow a\).

Step by step solution

01

Recognize the Types of Monotonic Sequences

First, analyze if the given monotonic sequence is increasing or decreasing.
02

Case I: Increasing Sequence

If the sequence is increasing, then \(a_{n+1} \geq a_n\) for all \(n \in \mathbb{N}\).
03

Use the Subsequence Limit

With an increasing sequence and given that \(a_{n_k} \rightarrow a\), we can use the definition of convergence: for any \(\epsilon > 0\), there exists \(K \in \mathbb{N}\) such that \(|a_{n_k} - a| < \epsilon\) for all \(k \geq K\).
04

Utilize the Monotonic Property

As the sequence is increasing, for any \(n \geq n_K\), we have that \(a_n \geq a_{n_K}\). Therefore, \(a - a_n \leq a - a_{n_K} < \epsilon\).
05

Prove Convergence of the Increasing Sequence

Since we have \(|a_n - a| = a - a_n \leq a - a_{n_K} < \epsilon\) for all \(n \geq n_K\), the sequence \((a_n)\) converges to \(a\) in case of an increasing sequence.
06

Case II: Decreasing Sequence

If the sequence is decreasing, then \(a_{n+1} \leq a_n\) for all \(n \in \mathbb{N}\).
07

Employ the Subsequence Limit

Similar to the increasing case, with a decreasing sequence and given that \(a_{n_k} \rightarrow a\), we can use the definition of convergence: for any \(\epsilon > 0\), there exists \(K \in \mathbb{N}\) such that \(|a_{n_k} - a| < \epsilon\) for all \(k \geq K\).
08

Apply the Monotonic Property

As the sequence is decreasing, for any \(n \geq n_K\), we have that \(a_n \leq a_{n_K}\). Therefore, \(a_n - a \leq a_{n_K} - a < \epsilon\).
09

Prove Convergence of the Decreasing Sequence

Since we have \(|a_n - a| = a_n - a \leq a_{n_K} - a < \epsilon\) for all \(n \geq n_K\), the sequence \((a_n)\) converges to \(a\) in case of a decreasing sequence. In both cases, we have shown that if a monotonic sequence has a convergent subsequence, then the whole sequence converges to the same limit.

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Key Concepts

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

Monotonic Sequences
Sequences that either always increase or decrease are called monotonic sequences.
They are important because they simplify the study of convergence. In mathematical terms:
  • An increasing sequence satisfies: \(a_{n+1} \geq a_n\) for all \(n \in \mathbb{N}\)
  • A decreasing sequence satisfies: \(a_{n+1} \leq a_n\) for all \(n \in \mathbb{N}\)

When studying the convergence of monotonic sequences, knowing that a sequence is either increasing or decreasing helps us apply different conclusions about their behavior. This is because in many cases, monotonic sequences that are bounded converge, which is a useful property in analysis when these sequences are encountered.
For example, if we have a bounded increasing sequence, it will eventually get closer to some upper limit but never go beyond it, which indicates convergence.
Subsequence Limits
A subsequence is a derived sequence that we obtain by selecting specific terms from the original sequence.
Subsequence limits are crucial in analyzing the behavior of sequences, particularly in determining if an entire sequence converges. A fundamental rule in analysis is that if a subsequence of a sequence converges to a certain limit, under specific conditions, it indicates that the original sequence might also converge to this limit.
As given in the exercise example, the sequence \((a_{n})\) is monotonic with a convergent subsequence \((a_{n_k})\) converging to \(a\). The convergence of the subsequence implies the original sequence trends toward \(a\) too. It's a great tool for checking convergence without testing every element of the sequence and is especially effective when combined with the known properties of monotonic sequences.
Sequence Convergence
Convergence describes a sequence whose terms approach a specific value as they progress towards infinity.
Formally, a sequence \((a_{n})\) converges to a limit \(a\) if, for every \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(n \geq N\), \(|a_{n} - a| < \epsilon\).
This definition essentially assures us that the points of the sequence eventually become arbitrarily close to \(a\).
In the context of monotonic sequences with a convergent subsequence, the convergence of the subsequence assists in proving that the entire sequence converges to the same limit. It鈥檚 insightful because it simplifies the concrete verification of convergence by examining just a part of the sequence. Verifying the limits of subsequences often involves easier calculations when compared to the entire sequence.
Real Analysis
Real Analysis is a fundamental branch of mathematics that explores the properties of real numbers, sequences, series, and functions.
It鈥檚 deeply concerned with the rigor in proving mathematical theories involving limits and convergence.
  • Real analysis provides tools such as the concept of limits, continuity, and convergence, which are essential for thoroughly understanding sequences and their behaviors.
  • In dealing with sequences, it gives us the theoretical foundation for monotonic sequences, bounded sequences, and their convergence properties.

Applying these concepts, as illustrated in the example exercise, is crucial for solving more complex problems involving sequence convergence. The solid, logical foundations laid by real analysis support a reliable understanding of the subtleties involved in series and continuous functions, giving students a clearer narrative when handling seemingly tricky sequence problems.

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Most popular questions from this chapter

Suppose \(\alpha\) and \(\beta\) are nonnegative real numbers. Let $$ a_{1}:=\alpha \quad \text { and } \quad a_{n+1}:=\beta+\sqrt{a_{n}} \quad \text { for } n \in \mathbb{N} $$ Show that \(\left(a_{n}\right)\) is convergent. Further, if \(a:=\lim _{n \rightarrow \infty} a_{n}\), then show that \(a=0\) if \(\alpha=0=\beta\), and \(a=(1+2 \beta+\sqrt{1+4 \beta}) / 2\) otherwise. (Hint: Consider the cases \(\sqrt{\alpha}+\beta \leq \alpha\) and \(\sqrt{\alpha}+\beta>\alpha\).)

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{R}\). (i) Assume that \(\left(a_{n}\right)\) is bounded above and \(a_{n} \not \rightarrow-\infty\). Define \(M_{n}:=\sup \left\\{a_{n}, a_{n+1}, \ldots\right\\}\) for \(n \in \mathbb{N} \quad\) and \(\quad M:=\inf \left\\{M_{1}, M_{2}, \ldots\right\\}\) Show that the sequence \(\left(M_{n}\right)\) converges to \(M\) and \(M\) is the largest cluster point of \(\left(a_{n}\right)\). (ii) Assume that \(\left(a_{n}\right)\) is bounded below and \(a_{n} \not \leftrightarrow \infty\). Define \(m_{n}:=\inf \left\\{a_{n}, a_{n+1}, \ldots\right\\}\) for \(n \in \mathbb{N} \quad\) and \(\quad m:=\sup \left\\{m_{1}, m_{2}, \ldots\right\\}\) Show that the sequence \(\left(m_{n}\right)\) converges to \(m\) and \(m\) is the smallest cluster point of \(\left(a_{n}\right)\). [See Exercise 16 for the definition of a cluster point.]

Given any \(x \in \mathbb{R}\), show that there is a nonnegative integer \(k\) and integers \(a_{k}, a_{k-1}, \ldots, a_{1}, a_{0}, b_{1}, b_{2}, \ldots\) between 0 and 9 such that $$ x=\pm \lim _{n \rightarrow \infty}\left(a_{k} 10^{k}+a_{k-1} 10^{k-1}+\cdots+a_{0}+\frac{b_{1}}{10}+\frac{b_{2}}{10^{2}}+\cdots+\frac{b_{n}}{10^{n}}\right) . $$ (Hint: If \(|x|<1\), set \(k=0=a_{0}\) and apply Exercise 26 to \(y:=|x|\), whereas if \(|x| \geq 1\), apply Exercise 27 to \(n:=[|x|]\) and Exercise 26 to \(y:=|x|-n .)\) [Note: It is customary to call \(a_{k}, a_{k-1}, \ldots, a_{0}, b_{1}, b_{2}, \ldots\) the digits of \(x\) and write the above expression for \(x\) as \(x=\pm a_{k} a_{k-1} \ldots a_{0} \cdot b_{1} b_{2} \ldots\), and call it the decimal expansion of \(x\).

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{R}\). Define the limit superior (or the upper limit) of \(\left(a_{n}\right)\) by $$ \limsup _{n \rightarrow \infty} a_{n}:=\left\\{\begin{array}{ll} \lim _{n \rightarrow \infty} M_{n} & \text { if }\left(a_{n}\right) \text { is bounded above and } a_{n} \not \rightarrow-\infty \\ \infty & \text { if }\left(a_{n}\right) \text { is not bounded above } \\ -\infty & \text { if } a_{n} \rightarrow-\infty \end{array}\right. $$ where the sequence \(\left(M_{n}\right)\) is as defined in Exercise 35 . Similarly, define the limit inferior (or the lower limit) of \(\left(a_{n}\right)\) by $$ \liminf _{n \rightarrow \infty} a_{n}:=\left\\{\begin{array}{ll} \lim _{n \rightarrow \infty} m_{n} & \text { if }\left(a_{n}\right) \text { is bounded below and } a_{n} \rightarrow \infty \\ -\infty & \text { if }\left(a_{n}\right) \text { is not bounded below, } \\ \infty & \text { if } a_{n} \rightarrow \infty \end{array}\right. $$ where the sequence \(\left(m_{n}\right)\) is as defined in Exercise 35 . If \(\left(a_{n}\right)\) is bounded, then show that the set \(C\) of all cluster points of \(\left(a_{n}\right)\) is nonempty, and moreover, $$ \limsup _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \sup \left\\{a_{n}, a_{n+1}, \ldots\right\\}=\max C $$ and $$ \liminf _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \inf \left\\{a_{n}, a_{n+1}, \ldots\right\\}=\min C. $$

For \(n \in \mathbb{N}\), let $$ a_{n}:=\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n} \quad \text { and } \quad b_{n}:=\frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{2 n-1} $$ Show that \(a_{n} \rightarrow \infty\) and \(b_{n} \rightarrow \infty\). (Hint: Example 2.10 (iii).)
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