Question

Assume that \(\bar{s}, \underline{s}\) ( or \(s\) ), \(\bar{t},\) and \(\underline{t}\) are in the extended reals, and show that the given inequalities or equations hold whenever their right sides are defined (not indeterminate). (a) \(\varlimsup_{n \rightarrow \infty}\left(-s_{n}\right)=-\underline{s}\) (b) \(\lim _{n \rightarrow \infty}\left(-s_{n}\right)=-\bar{s}\)

Short answer

Question: Prove the following statements for sequences of extended real numbers: (a) \(\varlimsup_{n \rightarrow \infty}(-s_n) = -\underline{s}\) (b) If \(s_n \rightarrow s\) as \(n \rightarrow \infty\), then \(-s_n \rightarrow -s\) Answer: (a) By understanding the definitions of lim sup, lim inf, and limits, we applied the property that the negation of infimum is equal to the supremum of the negation and vice versa to show that \(\varlimsup_{n \rightarrow \infty}(-s_n) = -\underline{s}\). (b) We first showed that for a sequence \((s_n)\) to have a limit, both its lim sup and lim inf must be equal. Then, using a similar approach as in statement (a), we proved that if \(s_n \rightarrow s\) as \(n \rightarrow \infty\), then \(-s_n \rightarrow -s\).

Step-by-step solution

Understand the Definitions

It is important to understand the definitions of lim sup, lim inf, and limits to prove the statements. The lim sup of a sequence \((s_n)\) is defined as \(\varlimsup_{n \rightarrow \infty} s_n = \inf_{n \in \mathbb{N}} \sup_{m \geq n} s_m\). The lim inf of a sequence \((s_n)\) is defined as \(\underline{s} = \sup_{n \in \mathbb{N}} \inf_{m \geq n} s_m\). Finally, limits are used to describe the value a sequence approaches in the infinite position, which we can denote as \(\lim_{n \rightarrow \infty} s_n = s\) for a sequence \((s_n)\).

Prove Statement (a)

We need to prove that \(\varlimsup_{n \rightarrow \infty}(-s_n) = -\underline{s}\). We can start by applying the definition of lim inf to the negation of the sequence elements. \(\varlimsup_{n \rightarrow \infty}(-s_n) = \inf_{n \in \mathbb{N}} \sup_{m \geq n}(-s_m)\). Now, we can use the property that the negation of infimum is equal to the supremum of the negation and vice versa. \(\inf_{n \in \mathbb{N}} \sup_{m \geq n}(-s_m) = -\sup_{n \in \mathbb{N}} \inf_{m \geq n} s_m = -\underline{s}\). By applying the definition of lim inf, we have proven statement (a), that \(\varlimsup_{n \rightarrow \infty}(-s_n) = -\underline{s}\).

Prove Statement (b)

We need to prove that \(\lim_{n \rightarrow \infty}(-s_n) = -\bar{s}\). For a sequence to have a limit, the lim sup and lim inf must be equal. If that is the case, we can apply a similar method to prove this statement as we did in proving statement (a). Since both the lim sup and lim inf are equal, denoted as \(\bar{s}\), it holds that \(\varlimsup_{n \rightarrow \infty} s_n = \underline{s} = \bar{s}\). We need to show that \(\lim_{n \rightarrow \infty}(-s_n) = -\bar{s}\). First, we find the negation of the lim sup of the sequence: \(\varlimsup_{n \rightarrow \infty}(-s_n) = -\bar{s}\). Next, we can apply the property that the negation of infimum is equal to the supremum of the negation and vice versa, as we did in proving statement (a). \(-\underline{s} = -\bar{s}\). Finally, we know that the limit of a sequence is the same value for both lim sup and lim inf. Since we have proven the equality of the negation of lim inf to the negation of lim sup, it follows that \(\lim_{n \rightarrow \infty}(-s_n) = -\bar{s}\), proving statement (b).

Key concepts

lim sup and lim inf

When dealing with sequences in real analysis, understanding the concepts of lim sup (limit superior) and lim inf (limit inferior) is crucial.

• The lim sup of a sequence \((s_n)\) is an upper bound on the limit of sequence elements. It reflects the 'tendency' of the largest values within segments of the sequence, in the long run. Mathematically, we define it as: \[\varlimsup_{n \rightarrow \infty} s_n = \inf_{n \in \mathbb{N}} \sup_{m \geq n} s_m\]
• Conversely, the lim inf describes a lower bound for these elements. It highlights the 'tendency' towards the smallest values in the sequence. This is given by:\[\underline{s} = \sup_{n \in \mathbb{N}} \inf_{m \geq n} s_m\]

Both give us a more nuanced view of a sequence's behavior, especially when simply determining the limit is insufficient or doesn't exist. In applications like the one from the exercise, switching signs means that the lim sup of the negated sequence equals the negated lim inf of the original sequence.

sequence convergence

A sequence is said to converge if it approaches a finite value as the index heads towards infinity. Convergence is a vital concept in understanding the behavior and properties of sequences. It tells us that, after a certain point, the values of the sequence get arbitrarily close to one another and to the limit.For a sequence \((s_n)\) to converge, both its lim sup and lim inf must meet at a single point. This is because:

• The lim sup representing the highest boundary, and
• The lim inf representing the lowest boundary.

When both align, they encapsulate the sequence tightly, indicating that all terms ultimately stabilize at this shared value. This is the essence behind proving statement (b) in the exercise, where the equality of lim sup and lim inf (\(\bar{s} = \underline{s}\)) allowed us to deduce the convergence of the sequence \((-s_n)\) to \(-\bar{s}\).

limit of a sequence

The limit of a sequence is the specific value that the sequence's terms converge to as the index approaches infinity. This concept helps us predict the eventual outcome of a sequence, or more formally, provides us with the "long-term" behavior of sequences.When a sequence \((s_n)\) has a limit, expressed as \(\lim_{n \rightarrow \infty} s_n = L\), it follows that the distance between \(s_n\) and \(L\) becomes negligible for large \(n\). This is rooted in the formal definition: for every \(\varepsilon > 0\), there exists a natural number \(N\) such that all sequence terms, beyond a certain index \(N\), satisfy \(|s_n - L| < \varepsilon\).This understanding was essential in proving statement (b) from the exercise. By affirming the equal values of lim sup and lim inf, the sequence \((-s_n)\) stabilizes around \(-\bar{s}\), fulfilling the formal requirement of a convergence towards this particular limit.