Question

Prove: If \(\lim _{n \rightarrow \infty} s_{n}=s\) and \(\left\\{s_{n}\right\\}\) has a subsequence \(\left\\{s_{n_{k}}\right\\}\) such that \((-1)^{k} s_{n_{k}} \geq 0\) then \(s=0\).

Short answer

Question: Prove that if the limit of a sequence \(s_n\) is \(s\) as \(n \rightarrow \infty\) and there is a subsequence \(s_{n_k}\) that satisfies \((-1)^k s_{n_k} \geq 0\), then \(s=0\). Answer: To prove this, we split the subsequence into two parts: even-indexed and odd-indexed terms. For the even-indexed subsequence terms, the limit is non-negative, which implies \(s \geq 0\). For the odd-indexed subsequence terms, the limit is non-positive, which implies \(s \leq 0\). Based on these results, we conclude that \(s = 0\).

Step-by-step solution

Recall the properties of subsequences and limits

Subsequences inherit the limit of the original sequence. In this case, if \(\lim_{n\rightarrow \infty} s_n = s\), then \(\lim_{k\rightarrow \infty} s_{n_k} = s\).

Evaluate the limit of the subsequence terms with different signs

Since \((-1)^k s_{n_k} \geq 0\), for even \(k=2m\), \(s_{n_{2m}} \geq 0\). For odd \(k=2m+1\), \(s_{n_{2m+1}} \leq 0\).

Split the subsequence into two parts: Even and odd

Based on the above observation, split the subsequence \(s_{n_k}\) into two parts: - Subsequence of terms with even indices, denoted as \({s_{n_{2m}}}\), where \(s_{n_{2m}} \geq 0\). - Subsequence of terms with odd indices, denoted as \({s_{n_{2m+1}}}\), where \(s_{n_{2m+1}} \leq 0\).

Evaluate the limit of the even-indexed subsequence

Since the limit of \(s_{n_k}\) as \(k\rightarrow \infty\) is equal to the limit of \(s_n\) as \(n\rightarrow \infty\), we know that \(\lim_{m\rightarrow \infty} s_{n_{2m}} = s\). Also, each term of this subsequence is non-negative \(s_{n_{2m}} \geq 0\). This implies that the limit \(s \geq 0\).

Evaluate the limit of the odd-indexed subsequence

Similarly, since the limit of \(s_{n_k}\) as \(k\rightarrow \infty\) is equal to the limit of \(s_n\) as \(n\rightarrow \infty\), we know that \(\lim_{m\rightarrow \infty} s_{n_{2m+1}} = s\). Also, each term of this subsequence is non-positive \(s_{n_{2m+1}} \leq 0\). This implies that the limit \(s \leq 0\).

Conclude that the limit \(s=0\)

Combining the results from steps 4 and 5, we see that \(s \geq 0\) and \(s \leq 0\). Therefore, we conclude that \(s=0\).

Key concepts

Limit of a Sequence

Understanding the limit of a sequence is fundamental in real analysis, as it helps to discern the behavior of sequences as their terms progress towards infinity. A limit of a sequence is a value that the terms of the sequence approach as the index increases indefinitely. Formally, we write this as \( \lim_{n\rightarrow\infty} a_n = L \) if for every positive number \( \epsilon \), there exists a natural number \( N \) such that for all \( n > N \) it holds that \( |a_n - L| < \epsilon \). The value \( L \) is called the limit, and the sequence is said to converge to \( L \).

For instance, in the given exercise, we are told that a sequence \( \{s_n\} \) converges to some limit \( s \). That is to say, as \( n \) grows larger, the terms of the sequence \( s_n \) are getting arbitrarily close to \( s \). This concept of a limit is intrinsic to the understanding of continuity, differentiation, and integration within calculus.

Subsequences and Convergence

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. For example, if we have a sequence \( \{a_n\} \) such as \( \{1,2,3,4,5,...\} \) then \( \{1,3,5,...\} \) is a subsequence. When we investigate the convergence of subsequences, we are essentially exploring whether these derived sequences adhere to the same limiting behavior as the original sequence.

In the given exercise, we look at a particular subsequence \( \{s_{n_k}\} \) where the terms alternate signs. The terms of this subsequence still converge to the same limit as the original sequence \( \{s_n\} \) owing to the properties of limits. The convergence of both the even and odd subsequences to the same limit \( s \) is crucial in deducing the limit's value, especially when contrasted with additional conditions such as sign alternation.

Properties of Limits

The properties of limits in real analysis are essential rules that allow us to manipulate limits in various ways while maintaining their fundamental characteristics. Some key properties include the limit of a constant sequence is the constant itself, the limit of the sum of two sequences is the sum of the limits, and if a sequence converges to a limit \( L \) then any subsequence also converges to \( L \).

In our problem, by employing these properties, we understand that irrespective of the alternating signs, if the original sequence \( \{s_n\} \) has a limit \( s \), then both even-indexed and odd-indexed subsequences inherited from \( \{s_n\} \) must converge to \( s \) as well. Since we have two subsequences bounded by non-negative and non-positive values respectively, and both sharing the same limit, this limit must then be zero. This crucial step harnesses the squeeze theorem, which states that if a sequence is squeezed between two sequences converging to the same limit, the sequence itself must converge to that same limit.