Problem 40 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=(-1)^{n}\left(1-\frac{1}{n}\right) $$

Step-by-Step Solution

Answer

The sequence \(a_{n} = (-1)^n \left(1-\frac{1}{n}\right)\) diverges.

Step 1: Identify the Sequence

The given sequence is \( a_n = (-1)^n \left(1 - \frac{1}{n}\right) \). Let's analyze the behavior as \( n \to \infty \).

Step 2: Determine Convergence or Divergence

Notice that \( (-1)^n \) makes the sequence alternate between positive and negative values. Since \( \lim_{n \to \infty} \left(1 - \frac{1}{n}\right) = 1 \), the product with \((-1)^n\) will make the sequence oscillate between values close to 1 and -1.

Step 3: Convergence Analysis

For the sequence to converge, it must approach a single fixed value. However, \((-1)^n \) causes the sequence to oscillate between values close to 1 and -1 indefinitely. Hence, it cannot settle at a single limit.

Step 4: Conclusion

As the oscillating behavior between two different values (1 and -1) persists, the sequence does not converge.

Key Concepts

Alternating Sequences

An alternating sequence is a sequence whose terms alternate in sign. In simple terms, this means that the terms switch back and forth between positive and negative values. For example, the sequence \(a_n = (-1)^n \) alternates because for even values of \( n \), \((-1)^n\) is positive, and for odd values, it is negative. Alternating sequences are interesting because their oscillating nature can affect whether they converge or diverge.

The sequence \( a_n = (-1)^n \left(1 - \frac{1}{n}\right) \) given in the exercise is an alternating sequence. As \( n \) increases, note how \((-1)^n\) influences each term to alternate between positive and negative. This alternating pattern means the sequence tries to oscillate rather than settle down to a single limiting value. Such behavior is a key attribute when analyzing the convergence or divergence of a sequence.

Limit of a Sequence

The limit of a sequence is the value that the terms of the sequence approach as the index \( n \) becomes very large. To determine whether a sequence converges is to check if such a limit exists. If all terms get arbitrarily close to a particular number as \( n \) tends towards infinity, the sequence converges to that number, which is called its limit.

In analyzing \( a_n = (-1)^n \left(1 - \frac{1}{n}\right) \), we separately look at the two parts. The expression \(1 - \frac{1}{n}\) approaches \(1\) as \( n \to \infty \), suggesting that if the sequence were not alternating, it could converge towards \(1\). Yet, the impact of \((-1)^n\) disrupts this approach by changing the sign alternately, preventing the sequence from settling at a particular constant value. Thus, despite \(1 - \frac{1}{n}\) having a well-defined limit, the whole sequence, when combined with \((-1)^n\), fails to converge to any single limit.

Divergence of Sequences

A sequence diverges when it does not converge, meaning there isn't a single, fixed value that the terms of the sequence approach. In other words, as the index \( n \) increases indefinitely, the sequence doesn't settle at any particular value, but rather continues to fluctuate or grow without bound.

The sequence \( a_n = (-1)^n \left(1 - \frac{1}{n}\right) \) exhibits divergence. The component \((-1)^n\) is the cause here, as it results in an unset or oscillating behavior. As such, instead of approaching one fixed limit, the sequence terms swing back and forth between values close to 1 and \(-1\), depending on whether \( n \) is even or odd. This persistent oscillation signifies that the sequence neither stabilizes nor approaches a specific value, which characterizes divergence in sequences. Thus, this sequence does not exhibit convergence, as it cannot be pinned down to a single limit.