Question

Let \(s_{n}=(-1)^{n}(1+1 / n)\). Show that \(\lim _{k \rightarrow \infty} s_{n_{k}}=1\) if and only if \(n_{k}\) is even for large \(k, \lim _{k \rightarrow \infty} s_{n_{k}}=-1\) if and only if \(n_{k}\) is odd for large \(k,\) and \(\left\\{s_{n_{k}}\right\\}\) diverges otherwise.

Short answer

Answer: The possible limits of the sequence \(s_{n_k}\) are 1 if \(n_k\) is even for large \(k\), -1 if \(n_k\) is odd for large \(k\), and the sequence diverges otherwise.

Step-by-step solution

Scenario 1: \(n_k\) is even for large \(k\)

If \(n_k\) is even for large \(k\), we can write it as \(n_k = 2m\) for some integer \(m\). We want to find \(\lim_{k\to\infty} s_{n_k}\) in this case. We can express the term \(s_{n_k}\) as \(s_{2m} = (-1)^{2m}(1+\frac{1}{2m})\). Since \((-1)^{2m} = 1\), we have \(s_{2m} = 1+\frac{1}{2m}\). Now, let's find the limit as \(k\to\infty\): $$\lim_{k\to\infty} s_{n_k} =\lim_{m\to\infty} s_{2m} = \lim_{m\to\infty} \left(1+\frac{1}{2m}\right).$$ As \(m\) tends to infinity, \(\frac{1}{2m}\) tends to 0, so the limit is \(1+0 = 1\).

Scenario 2: \(n_k\) is odd for large \(k\)

If \(n_k\) is odd for large \(k\), we can write it as \(n_k = 2m + 1\) for some integer \(m\). We want to find \(\lim_{k\to\infty} s_{n_k}\) in this case. We can express the term \(s_{n_k}\) as \(s_{2m+1} = (-1)^{2m+1}(1+\frac{1}{2m+1})\). Since \((-1)^{2m+1} = -1\), we have \(s_{2m+1} = -\left(1+\frac{1}{2m+1}\right)\). Now, let's find the limit as \(k\to\infty\): $$\lim_{k\to\infty} s_{n_k} =\lim_{m\to\infty} s_{2m+1} = \lim_{m\to\infty} \left(-\left(1+\frac{1}{2m+1}\right)\right).$$ As \(m\) tends to infinity, \(\frac{1}{2m+1}\) tends to 0, so the limit is \(-1(1+0) = -1\).

Scenario 3: General case

If the sequence \(n_k\) does not converge to even or odd indices for large \(k\), we cannot determine the limit of \(s_{n_k}\) because it alternates between the two limits found in the previous scenarios: 1 and -1. In this case, we can say that the sequence \(\{s_{n_k}\}\) diverges, as it oscillates between 1 and -1 and does not converge to a unique limit. Finally, we can conclude that \(\lim_{k\to\infty} s_{n_k} = 1\) if and only if \(n_k\) is even for large \(k\), \(\lim_{k\to\infty} s_{n_k} = -1\) if and only if \(n_k\) is odd for large \(k\), and the sequence \(\{s_{n_k}\}\) diverges otherwise.

Key concepts

Sequences and Limits

In real analysis, the concept of sequences and limits is fundamental to understanding how numbers behave as they approach infinity. A sequence is simply an ordered list of numbers. The notation \( \{s_n\} \) represents a sequence where \( n \) indicates the position of each term. Each term in the sequence is defined by a specific rule or formula, like \( s_n = (-1)^n(1+\frac{1}{n}) \) in this exercise.

The limit of a sequence \( \{s_n\} \) is the value that the terms of the sequence get closer to as \( n \) becomes very large. If the terms of the sequence approach a specific number, the sequence is said to converge, and this specific number is called the limit. For example, if the sequence approaches the number 1 as \( n \) increases, its limit is 1.

But not all sequences have a limit. If the terms of the sequence do not approach any particular value, the sequence is said to diverge. This means that the sequence does not settle down to a single number as \( n \) increases without bound.

Convergence and Divergence

Convergence and divergence are two essential concepts when studying sequences and series in real analysis. A sequence converges if its terms approach a definite value as the sequence progresses towards infinity. In mathematical terms, a sequence \( \{a_n\} \) converges to a limit \( L \) if, for every positive number \( \epsilon \), there is a point in the sequence beyond which all terms are within \( \epsilon \) of \( L \).

In the exercise, if \( n_k \) is even, the subsequence \( \{s_{n_k}\} \) converges to 1; if \( n_k \) is odd, it converges to -1, due to alternating signs based on parity. However, if the sequence does not consistently follow even or odd indices, it fails to approach a definitive limit, instead alternating between values. This behavior is characteristic of a diverging sequence. Divergence implies that the sequence lacks a well-defined limit and instead oscillates indefinitely without settling down.

Understanding convergence and divergence helps in determining the stability and predictability of sequences, which is critical in advanced mathematical analysis and its applications.

Alternating Series

An alternating series is a series in which the terms are alternately positive and negative. This kind of sequence often emerges in real analysis and can have various behaviors based on its construction. The sequence you are dealing with in this exercise is an excellent example of an alternating sequence, as defined by \( (-1)^n(1+\frac{1}{n}) \).

The alternating series takes the form of \( (-1)^n \) which switches the sign of each term based on whether \( n \) is even or odd. For even \( n \), \((-1)^n \) equals 1, keeping the term positive, while for odd \( n \), \((-1)^n \) equals -1, making the term negative.

Working with alternating sequences requires understanding the addition or subtraction properties of alternating terms, especially in how they affect convergence. In this specific example, the convergence or divergence of the sequence \( \{s_{n_k}\} \) relies heavily on the parity of \( n_k \). If the parity remains consistent (all even or all odd), the sequence converges to a limit. If the parity alternates without some regularity, the sequence diverges.

Thus, alternating series and sequences are a crucial part of understanding how sequences behave and what the convergence criteria might be in complex series.