Question

Find \(\bar{s}\) and \(\underline{s}\) (a) \(s_{n}=\left[(-1)^{n}+1\right] n^{2}\) (b) \(s_{n}=\left(1-r^{n}\right) \sin \frac{n \pi}{2}\) (c) \(s_{n}=\frac{r^{2 n}}{1+r^{n}} \quad(r \neq-1)\) (d) \(s_{n}=n^{2}-n\) (e) \(s_{n}=(-1)^{n} t_{n}\) where \(\lim _{n \rightarrow \infty} t_{n}=t\)

Short answer

a) \(s_n = \left[(-1)^{n}+1\right] n^{2}\) b) \(s_n = \left(1-r^{n}\right) \sin \frac{n \pi}{2}\) with \(|r| < 1\) c) \(s_n =\frac{r^{2 n}}{1+r^{n}} \quad(r \neq-1)\) d) \(s_n = n^2 - n\) e) \(s_n = (-1)^{n} t_{n}\), where \(\lim_{n \rightarrow \infty} t_{n} = t\) Answer: a) \(\bar{s}=+\infty\) and \(\underline{s}=0\) b) \(\bar{s}=1\) and \(\underline{s}=-1\) c) \(\bar{s}=1\) and \(\underline{s}=0\) d) \(\bar{s}=+\infty\) and \(\underline{s}=0\) e) \(\bar{s}=t\) and \(\underline{s}=-t\)

Step-by-step solution

Examine the sequence

Observe that \(s_n = \left[(-1)^{n}+1\right] n^{2}\). Consider the two possibilities for \(n\): odd and even.

Analyze the odd terms

When \(n\) is odd, \((-1)^n = -1\). Thus, for odd \(n\), the sequence becomes \(s_n = 0 \cdot n^2 = 0\). The accumulation points here are {0}.

Analyze the even terms

When \(n\) is even, \((-1)^n = 1\). Thus, for even \(n\), the sequence becomes \(s_n = 2 \cdot n^2\). Clearly, these values of \(s_n\) are increasing as \(n\) increases, so the accumulation points are unbounded as \(n \rightarrow \infty\).

Calculate superior and inferior limits

The inferior limit \(\underline{s}\) is the supremum of all the accumulation points for odd values of \(n\) which is 0. The superior limit \(\bar{s}\) is the supremum of all the accumulation points for even values of \(n\), which is \(\infty\). Thus, we have: \(\bar{s}=+\infty\) and \(\underline{s}=0\) #Case (b)#

Rewrite the sequence

Observe that \(s_n = \left(1-r^{n}\right) \sin \frac{n \pi}{2}\) can be rewritten as \(s_n = \left(1-r^{n}\right)\cdot (-1)^k\) where \(k\) is an integer and \(n = 4k + 1\) or \(n = 4k + 3\).

Find the supremum and infimum

Note that \(-1 \leq (-1)^k \leq 1\). We know that \(0 \leq 1 - r^n \leq 1\) since \(|r| < 1\). Therefore, \(-1(1 - r^n) \leq s_n \leq 1(1 - r^n)\). So, as \(n \rightarrow \infty\), the inferior limit \(\underline{s}\) is -1, and the superior limit \(\bar{s}\) is 1, thus: \(\bar{s}=1\) and \(\underline{s}=-1\) #Case (c)#

Analyze the sequence

Observe that \(s_n =\frac{r^{2 n}}{1+r^{n}} \quad(r \neq-1)\). Since \(r \neq -1\), we know that the sequence is always non-negative as the numerator is always positive and the denominator is never zero.

Find the supremum and infimum

As \(n\) goes to infinity, \(r^n\) goes to zero if \(|r|<1\) and goes to infinity if \(|r|>1\). In both cases, \(r^{2n}\) goes to zero as \(n \rightarrow \infty\). Thus, the inferior limit \(\underline{s}\) is 0. Since the sequence is non-negative and bounded by 1, the superior limit \(\bar{s}\) is 1. Thus, we have: \(\bar{s}=1\) and \(\underline{s}=0\) #Case (d)#

Analyze the sequence

Observe that \(s_n = n^2 - n\). The sequence is increasing as \(n \rightarrow \infty\), so the inferior limit of the sequence is \(0\).

Find the superior limit

As the sequence increases and has no upper bound, the superior limit is \(\infty\). Thus, we have: \(\bar{s}=+\infty\) and \(\underline{s}=0\) #Case (e)#

Analyze the sequence

Observe that \(s_n = (-1)^{n} t_{n}\) where \(\lim_{n \rightarrow \infty} t_{n} = t\).

Find the supremum and infimum

We know that \(t_n\) approaches \(t\) as \(n \rightarrow \infty\), but the sequence alternates signs. Therefore, the inferior limit \(\underline{s}\) is \(-t\) and the superior limit \(\bar{s}\) is \(t\) (assuming \(t \geq 0\); otherwise, swap the limits). Thus, we have: \(\bar{s}=t\) and \(\underline{s}=-t\)

Key concepts

Supremum and Infimum

In the realm of real analysis, understanding the concept of supremum and infimum is crucial for dealing with sequences and sets. These terms refer to the least upper bound and the greatest lower bound of a set, respectively.

• The supremum of a set is the smallest number that is greater than or equal to every number in the set. If the supremum is within the set, it's called the maximum.
• The infimum, on the other hand, is the largest number that is less than or equal to every number in the set. If the infimum is an element of the set, it is referred to as the minimum.

Knowing the supremum and infimum helps in determining the bounds of a sequence or set. For instance, if a sequence like \(s_n = (-1)^n n^2\) fluctuates between two values, its supremum and infimum can unveil the "envelope" within which the sequence fluctuates. In this sequence, depending on whether \(n\) is odd or even, the values shift between 0 and leaving it unbounded as \(n ightarrow \infty\). Therefore, knowing these values assists in deeper analysis.

Limit Superior and Inferior

When dealing with sequences, understanding the concepts of limit superior and limit inferior becomes essential, especially when sequences do not converge neatly to a single value.

• The limit superior, \( \bar{s} \), is the limit of the suprema of the tail ends of a sequence. It is the eventual "greatest height" the sequence approaches as we move infinitely far along the sequence.
• The limit inferior, \( \underline{s} \), represents the limit of the infima of the tails of the sequence. It describes the "lowest valley" the sequence touches in the long run.

This concept can help in analyzing the irregular sequences such as\(s_{ n } = ( - 1 )^n t_{ n }\). Here, as the terms oscillate, the limit inferior and superior will show us the bounds the sequence approaches infinitely. If \(t_{ n }\) converges to \(t\), then as we evaluate this sequence, we realize: the superior limit \(\bar{s} = t\)and inferior limit \(\underline{s} = -t\). These limits help uncover the long-term trends of oscillating or bounded sequences.

Real Analysis Sequences

Real analysis provides a rigorous foundation for dealing with "what happens after infinity" in sequences, which is the study of the limits. Sequences are essentially ordered lists of numbers. Their behavior as they progress to infinity can vary widely. In real analysis, understanding sequences is pivotal as it lays the groundwork for defining functions and can explain convergence and divergence concepts.

• Convergent sequences approach a single value as more terms are added. For example, such a sequence might be \(t_n = \frac{1}{n}\), which approaches 0 as \(n\) increases.
• Divergent sequences do not settle at any finite boundary. Consider \(s_n = n^2 - n\), which increases and heads towards infinity. Understanding its trend enables prediction of behaviors in other sequences with similar structures.
• Subsequences and oscillations can complicate the analysis,but exploring concepts like limit superior/inferior, supremum/infimum offer deeper insights.

In this context, a sound grasp of real analysis' employment of sequences can yield powerful techniques for resolving calculus problems and further understanding more advanced mathematical theorems.