Question

Find all numbers \(L\) in the extended reals that are limits of some subsequence of \(\left\\{s_{n}\right\\}\) and, for each such \(L,\) choose a subsequence \(\left\\{s_{n_{k}}\right\\}\) such that \(\lim _{k \rightarrow \infty} s_{n_{k}}=L\). (a) \(s_{n}=(-1)^{n} n\) (b) \(s_{n}=\left(1+\frac{1}{n}\right) \cos \frac{n \pi}{2}\) (c) \(s_{n}=\left(1-\frac{1}{n^{2}}\right) \sin \frac{n \pi}{2}\) (d) \(s_{n}=\frac{1}{n}\) (e) \(s_{n}=\left[(-1)^{n}+1\right] n^{2}\) (f) \(s_{n}=\frac{n+1}{n+2}\left(\sin \frac{n \pi}{4}+\cos \frac{n \pi}{4}\right)\)

Short answer

(a) \(s_n=(-1)^n n\) 1. \(\lim_{n\to\infty} 2n = +\infty\) 2. \(\lim_{n\to\infty} -(2n - 1) = -\infty\) (b) \(s_n = \left(1 + \frac{1}{n}\right)\cos\frac{n\pi}{2}\) 1. \(\lim_{n\to\infty} \left(1 + \frac{1}{4n}\right) = 1\) 2. \(\lim_{n\to\infty} 0 = 0\) 3. \(\lim_{n\to\infty} -\left(1+\frac{1}{4n-2}\right) = -1\) (c) \(s_n = \left(1-\frac{1}{n^2}\right)\sin\frac{n\pi}{2}\) 1. \(\lim_{n\to\infty} 0 = 0\) 2. \(\lim_{n\to\infty} \left(1-\frac{1}{(4n-1)^2}\right) = 1\) 3. \(\lim_{n\to\infty} -\left(1-\frac{1}{(4n-3)^2}\right) = -1\) (d) \(s_n = \frac{1}{n}\) 1. \(\lim_{n\to\infty} \frac{1}{n}=0\) (e) \(s_n = \left[(-1)^n +1 \right]n^2\) 1. \(\lim_{n\to\infty} 4n^2 = +\infty\) 2. \(\lim_{n\to\infty} 0=0\) (f) \(s_n=\frac{n+1}{n+2}\left(\sin\frac{n\pi}{4}+\cos\frac{n\pi}{4}\right)\) 1. \(\lim_{n\to\infty} \frac{n+1}{n+2}\left(1 + 1\right) = 1\) 2. \(\lim_{n\to\infty} \frac{n+1}{n+2}\left(1 - 1\right) = 0\) 3. \(\lim_{n\to\infty} \frac{n+1}{n+2}\left(-1 + 1\right) = 0\) 4. \(\lim_{n\to\infty} \frac{n+1}{n+2}\left(-1 - 1\right) = -1\)

Step-by-step solution

Determine the subsequences

For the sequence \(s_n=(-1)^n n\), we notice that there are two subsequences: 1. \(s_{2n} = (-1)^{2n}(2n) = 2n\) 2. \(s_{2n-1} = (-1)^{2n-1}(2n-1) = -(2n-1)\). (a)

Find the limits

The limits for these subsequences are: 1. \(\lim_{n\to\infty} 2n = +\infty\) 2. \(\lim_{n\to\infty} -(2n - 1) = -\infty\). (b)

Determine the subsequences

For the sequence \(s_n = \left(1 + \frac{1}{n}\right)\cos\frac{n\pi}{2}\), we can consider the four subsequences: 1. \(s_{4n}=\left(1+\frac{1}{4n}\right)\cos\frac{4n\pi}{2} = \left(1+\frac{1}{4n}\right)\cos (2n\pi) = 1+\frac{1}{4n}\) 2. \(s_{4n-1}=\left(1+\frac{1}{4n-1}\right)\cos\frac{(4n-1)\pi}{2} = 0.\) 3. \(s_{4n-2}=\left(1+\frac{1}{4n-2}\right)\cos\frac{(4n-2)\pi}{2} = -\left(1+\frac{1}{4n-2}\right)\) 4. \(s_{4n-3}=\left(1+\frac{1}{4n-3}\right)\cos\frac{(4n-3)\pi}{2} = 0\). (b)

Find the limits

The limits for these subsequences are: 1. \(\lim_{n\to\infty} \left(1 + \frac{1}{4n}\right) = 1\) 2. \(\lim_{n\to\infty} 0 = 0\) 3. \(\lim_{n\to\infty} -\left(1+\frac{1}{4n-2}\right) = -1\). (c)

Determine the subsequences

For the sequence \(s_n = \left(1-\frac{1}{n^2}\right)\sin\frac{n\pi}{2}\), we consider the following subsequences: 1. \(s_{4n} = 0\) 2. \(s_{4n-1} = \left(1-\frac{1}{(4n-1)^2}\right)\) 3. \(s_{4n-2} = 0\) 4. \(s_{4n-3} = -\left(1-\frac{1}{(4n-3)^2}\right)\) (c)

Find the limits

The limits for these subsequences are: 1. \(\lim_{n\to\infty} 0 = 0\) 2. \(\lim_{n\to\infty} \left(1-\frac{1}{(4n-1)^2}\right) = 1\) 3. \(\lim_{n\to\infty} -\left(1-\frac{1}{(4n-3)^2}\right) = -1\). (d)

Determine the subsequences

For the sequence \(s_n = \frac{1}{n}\), the only relevant subsequence is the sequence itself. (d)

Find the limit

The limit for this subsequence is: 1. \(\lim_{n\to\infty} \frac{1}{n}=0\). (e)

Determine the subsequences

For the sequence \(s_n = \left[(-1)^n +1 \right]n^2\), we have two subsequences: 1. \(s_{2n} = (1+1)(2n)^2 = 4n^2\) 2. \(s_{2n-1} = (1-1)((2n-1)^2)=0\) (e)

Find the limits

The limits for these subsequences are: 1. \(\lim_{n\to\infty} 4n^2 = +\infty\) 2. \(\lim_{n\to\infty} 0=0\). (f)

Determine the subsequences

For the sequence \(s_n=\frac{n+1}{n+2}\left(\sin\frac{n\pi}{4}+\cos\frac{n\pi}{4}\right)\), we will consider four subsequences where \(n \equiv 0, 1, 2, 3 \pmod{4}\), and analyze their behavior under each subsequence. (f)

Find the limits

Based on the trigonometric identities, the limits for the different sequences are: 1. \(\lim_{n\to\infty} \frac{n+1}{n+2}\left(1 + 1\right) = 1\) 2. \(\lim_{n\to\infty} \frac{n+1}{n+2}\left(1 - 1\right) = 0\) 3. \(\lim_{n\to\infty} \frac{n+1}{n+2}\left(-1 + 1\right) = 0\) 4. \(\lim_{n\to\infty} \frac{n+1}{n+2}\left(-1 - 1\right) = -1\).

Key concepts

Real Analysis

Real analysis is a fundamental area of mathematics that deals with properties of real numbers and real-valued sequences and functions. The heart of real analysis lies in understanding the concepts of convergence and limits, which are essential in studying continuity, integration, and differentiation on the real number line.

A sequence of real numbers is an ordered list \(s_1, s_2, s_3, \dots\), and these sequences can exhibit various behaviors as we move along their terms. They might approach a specific value, oscillate without settling down, or grow without bound. As such, real analysis provides the tools and theorems necessary to analyze and understand these behaviors, which is crucial not only for theoretical mathematics but also for applications in science and engineering.

Extended Reals

The extended real number system is an expansion of the real numbers to include two 'infinite' elements: \(+\infty\) and \( -\infty\). This extension is crucial because it allows us to discuss limits and convergence of sequences when they grow without bound in the positive or negative direction.

For instance, when we examine sequences that become arbitrarily large and positive, we say they tend toward \(+\infty\), as we saw in the sequence \(s_{n}=(-1)^{n} n\). On the other hand, the sequence \(s_{n} = \frac{1}{n}\) gets arbitrarily close to zero, reflecting another aspect of limit behavior, which can also be properly captured within the extended reals. By incorporating these infinite elements, the extended reals complete the real line, enabling a comprehensive analysis of all types of convergence within sequences.

Limit of a Sequence

The limit of a sequence is the value that the elements of the sequence approach as the index increases indefinitely. In mathematical terms, we say the limit of \(s_n\) as \(n\) approaches infinity is \(L\) if, for any given positive number \(\epsilon\), there is a corresponding index \(N\) such that for all \(n > N\), the distance between \(s_n\) and \(L\) is less than \(\epsilon\). It's denoted as \(\lim_{n \to \infty} s_n = L\).

Importantly, the definition of a limit accommodates sequences that converge to a real number, as well as those that tend toward \(+\infty\) or \( -\infty\). Limits are fundamental in identifying the long-term behavior of sequences, regardless of their initial erratic behavior. Our textbook solutions reveal these limits through careful examination—for instance, the sequence \(s_n = \frac{1}{n}\) has a limit of 0 while the sequence \(s_n = (-1)^{n} n\) does not have a finite limit, but instead diverges to \(+\infty\) or \( -\infty\).

Convergence of Sequences

Convergence is a key notion when studying real analysis and limits. A sequence is said to converge to a limit \(L\) in the extended real numbers if, as \(n\) gets larger, the terms \(s_n\) get arbitrarily close to \(L\) and stay that close.

For a sequence to converge, it must satisfy two main criteria. First, the limit must be well-defined, meaning it is a specific number in the extended reals; second, the sequence's terms must approach this limit and not deviate away as \(n\) increases. With the boundaries of convergence in mind, we can understand why some sequences, such as \(s_{n} = \frac{1}{n}\), which tends to zero, are considered convergent, while others like \(s_{n}=(-1)^{n} n\), which increases without limit, are not.

Sequences can also have subsequences that converge to different limits, which was the focus of the original textbook exercise. Identifying these subsequences and understanding their convergence helps students deeply appreciate the complexity and the beauty of sequence behaviors in real analysis.