Question

Prove: if \(\lim _{n \rightarrow \infty} s_{n}=s\) and \(\lim _{n \rightarrow \infty} t_{n}=t,\) where \(s\) and \(t\) are in the extended reals, then $$ \lim _{n \rightarrow \infty} s_{n} t_{n}=s t $$ provided that \(s t\) is defined in the extended reals.

Short answer

Based on the given step-by-step solution, the short answer to the problem is that, in all cases, the product of converging sequences in extended real numbers converges to the product of their respective limits (\(\lim _{n \rightarrow \infty} s_{n} t_{n} = st\)). The solution covers various cases and their sub-cases, applying properties of limits and carefully addressing each scenario.

Step-by-step solution

Understanding the limit concept in extended real numbers

The extended real numbers consist of real numbers (\(\mathbb{R}\)) and two elements \(+\infty\) and \(-\infty\). For any sequence of extended real numbers, we can apply the properties of limits which hold in \(\mathbb{R}\), as well as handle cases specifically for \(+\infty\) and \(-\infty\).

Identifying and handling possible cases

To cover all possibilities, we need to split our analysis into various cases: Case 1: \(s\) and \(t\) are both real numbers (\(s, t\in \mathbb{R}\)) Case 2: One of \(s\) or \(t\) belongs to the extended real numbers (\(s,t\in \overline{\mathbb{R}}=\mathbb{R} \cup \{+\infty, -\infty\}\)), and the other is real Case 3: Both \(s\) and \(t\) belong to the extended real numbers (\(s, t\in \overline{\mathbb{R}}\))

Proving Case 1

When both \(s\) and \(t\) are real numbers, we can apply the usual properties of limits in \(\mathbb{R}\). Specifically, the product of the two sequences will also converge to the product of the respective limits: $$ \lim _{n \rightarrow \infty} s_{n} t_{n} = \left(\lim_{n\rightarrow \infty} s_n\right) \left(\lim_{n\rightarrow \infty} t_n\right) = st $$

Proving Case 2

In this case, we need to handle each sub-case separately: Sub-case 1: One of them, say \(s\), is \(+\infty\) (or \(-\infty\)), and \(t\) is a non-zero real number. Then the product \(st\) will be either \(+\infty\) or \(-\infty\), and similarly, the product sequence \(s_n t_n\) will converge to \(+\infty\) or \(-\infty\). Sub-case 2: One of them, say \(s\), is \(+\infty\) (or \(-\infty\)), and \(t=0\). In this case, the product \(st\) is not defined in the extended reals, therefore the given statement is vacuously true.

Proving Case 3

Finally, we handle the case when both \(s\) and \(t\) belong to the extended real numbers. There are four possibilities for the combinations of \(s\) and \(t\): 1. \(s=+\infty\), \(t=+\infty\): In this case, \(s_n t_n\) converges to \(+\infty\), and \(st = (+\infty)(+\infty) = +\infty\). 2. \(s=+\infty\), \(t=-\infty\): In this case, \(s_n t_n\) converges to \(-\infty\), and \(st = (+\infty)(-\infty) = -\infty\). 3. \(s=-\infty\), \(t=+\infty\): In this case, \(s_n t_n\) converges to \(-\infty\), and \(st = (-\infty)(+\infty) = -\infty\). 4. \(s=-\infty\), \(t=-\infty\): In this case, \(s_n t_n\) converges to \(+\infty\), and \(st = (-\infty)(-\infty) = +\infty\). Hence, we have shown that in each case, the product sequence \(s_n t_n\) converges to the product of the respective limits \(st\). This completes the proof.

Key concepts

Limits in Extended Reals

The concept of limits in extended real numbers expands upon the familiar territory of real analysis by including two additional elements: \(+\infty\) and \(-\infty\). This extension provides a more robust framework for studying limits, especially those of sequences that may not converge within the standard real number system.
In this extended realm, limits are handled slightly differently to accommodate these infinite values. For instance, a sequence \(a_n\) might grow without bound, and we say it converges to \(+\infty\). Similarly, if the terms of the sequence decrease without bound, the sequence converges to \(-\infty\).
This expanded definition allows us to consider otherwise undefined products or sums, provided they fall into cases where the arithmetic still makes sense, such as \((+\infty) \times 3 = +\infty\).

Product of Sequences

When dealing with sequences, the product of two sequences \((s_n)\) and \((t_n)\) is defined as a new sequence \((s_n t_n)\), where each term is the product of the corresponding terms in the original sequences.
In the context of the extended reals, this means that if \(s_n\) approaches a limit \(s\) and \(t_n\) approaches a limit \(t\), then the product sequence \(s_n t_n\) should naturally approach \(st\) provided \(st\) is well-defined in the extended reals.
A common hurdle in understanding this product is the handling of infinities. For instance:

• \((+\infty) \times x = +\infty\) for any positive non-zero \(x\).
• \((+\infty) \times 0\) is considered an indeterminate form.

Therefore, any proof or analysis must carefully consider these special infinite cases to ensure the product is valid.

Convergence of Sequences

The convergence of sequences is a foundational concept in calculus and analysis. A sequence \((a_n)\) is said to converge to a limit \(L\) if for every positive number \(\epsilon\), there exists an integer \(N\) such that for all \(n > N\), \(|a_n - L| < \epsilon\). In simple terms, the terms of the sequence get arbitrarily close to \(L\) as \(n\) becomes very large.
When extending this concept to include \(+\infty\) and \(-\infty\), a sequence \((a_n)\) converges to \(+\infty\) if for each positive number \(M\), there is an integer \(N\) such that for all \(n > N\), \(a_n > M\). A similar definition applies to \(-\infty\).
This threshold-based definition allows us to handle sequences that do not necessarily settle into a single real number. Despite their divergence from typical point-based convergence, these sequences still follow a form of convergence in their infinite approach.