Question

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $$\left\\{a_{n}\right\\}=\left\\{\frac{(-1)^{n+1}}{n}\right\\}$$

Short answer

The sequence converges with a limit of 0.

Step-by-step solution

Understand the Sequence

The sequence in question is given by \( \{a_n\} = \{(-1)^{n+1}/n\} \). This means that every term of the sequence alternates in sign because of the \((-1)^{n+1}\) factor, and each term magnitude is given by \(1/n\).

Analyze the Behavior of Terms

The terms of the sequence are: \( a_1 = 1, a_2 = -\frac{1}{2}, a_3 = \frac{1}{3}, a_4 = -\frac{1}{4} \), and so on. As \( n \) increases, the magnitude of \( a_n \) gets smaller because \( \frac{1}{n} \to 0 \).

Determine Convergence or Divergence

To determine convergence, evaluate the limit of \( a_n \) as \( n \to \infty \). We have \( \lim_{n \to \infty} \frac{(-1)^{n+1}}{n} = 0 \) since the absolute value of \( a_n \), \( \frac{1}{n} \to 0 \), and alternating signs do not affect the limit.

Conclusion

Since the limit of the sequence \( \lim_{n \to \infty} a_n = 0 \), the sequence converges, and its limit is 0.

Key concepts

Alternating Series

An alternating series is a sequence of numbers where the signs of the terms alternate between positive and negative. In the sequence \( \{ a_n \} = \{ (-1)^{n+1}/n \} \), the factor \((-1)^{n+1}\) guarantees this alternation. For instance, when \( n \) is even, the term is negative, and when \( n \) is odd, the term is positive. Alternating series have special rules for convergence that often depend on whether the magnitude of each successive term is decreasing and if those magnitudes approach zero.

In simple terms, the sequence keeps flipping directions like a pendulum. However, due to the decreasing magnitude from \(1/n\), the swings are smaller and smaller, which influences the convergence of the sequence.

Limits of Sequences

A limit of a sequence is a value that the terms of the sequence approach as the number of terms goes to infinity. In mathematical terms, if \( \lim_{n \to \infty} a_n = L \), then \( L \) is the limit. For the sequence \( \left\{ (-1)^{n+1}/n \right\} \), we evaluate whether a limit exists by observing the behavior of the sequence as \( n \) becomes very large.

The terms in this sequence get smaller in magnitude because \(1/n\) becomes tiny, approaching zero as \(n\) becomes infinitely large. Therefore, the limit of this particular sequence is 0 because the alternating sign does not prevent the term \( \frac{1}{n} \) from shrinking towards zero.

Sequence Analysis

Sequence analysis involves examining the pattern and behavior of the terms in a sequence. When analyzing \(\{ a_n \} = \{ (-1)^{n+1}/n \}\), it helps to generate the initial few terms. This provides insight into the sequence's nature. For instance, the initial terms are \( a_1 = 1, a_2 = -1/2, a_3 = 1/3, a_4 = -1/4 \).

Through analysis, we observe that each term alternates in sign and diminishes in magnitude. This alternating pattern and reduction in term size hint towards a potential convergence point. Augmenting this analysis with formal limit evaluation can fully affirm the sequence’s behavior at infinity.

Convergent Sequences

A convergent sequence is one where the terms approach a specific value, known as the limit, as \( n \) goes to infinity. For \( \left\{ (-1)^{n+1}/n \right\} \), we validated convergence by showing \( \lim_{n \to \infty} \frac{(-1)^{n+1}}{n} = 0 \). This demonstrates that although the sequence alternates, the terms become negligible as they move closer to zero.

Understanding convergence is crucial in mathematics because it tells whether a sequence stabilizes to a certain value or not. In practical terms, a convergent sequence behaves predictably in the long run. Once we're confident that the sequence converges, we can further use this information in more complex mathematical analyses or applications.