Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(-1, \frac{2}{3},-\frac{3}{5}, \frac{4}{7},-\frac{5}{9}, \ldots\)

Step-by-Step Solution

Answer

The sequence is given by \(a_n = \frac{(-1)^n n}{2n-1}\) and it converges to \(\frac{1}{2}\).

Step 1: Identify the Pattern

Observe the given sequence: \[-1, \frac{2}{3}, -\frac{3}{5}, \frac{4}{7}, -\frac{5}{9}, \ldots\]There seems to be a pattern with alternate signs and fractions.

Step 2: Establish Numerator Formula

Notice that the numerators are \(-1, 2, -3, 4, -5, \ldots\). They follow a pattern of \((-1)^n \cdot n\). Ensure that this pattern fits the sequence for \(n = 1, 2, 3, \ldots\).

Step 3: Establish Denominator Formula

Now observe the denominators \(1, 3, 5, 7, 9, \ldots\). These are odd numbers, which can be represented by the formula \((2n - 1)\).

Step 4: Combine to Form Explicit Formula

Combine these observations to form the explicit formula:\[a_n = \frac{(-1)^n \, n}{2n - 1}\]

Step 5: Determine Convergence or Divergence

Examine the behavior of the sequence as \(n\) approaches infinity. The alternation of signs does not affect convergence, only the size of \(a_n\). Because \(\frac{n}{2n-1}\) approaches \(\frac{1}{2}\) as \(n\) grows, the sequence converges to zero.

Step 6: Find the Limit

Calculate the limit:\[\lim_{n \to \infty} \frac{(-1)^n \, n}{2n - 1} \rightarrow \lim_{n \to \infty} \frac{n}{2n-1} = \frac{1}{2}\] Thus, the limit of the sequence as \(n \to \infty\) is \(\frac{1}{2}\).

Key Concepts

Explicit Formula

An explicit formula for a sequence is like a magical rulebook. It allows us to find any term in the sequence without listing all the previous ones. Imagine you have a sequence given by terms like -1, \(\frac{2}{3}\), -\(\frac{3}{5}\), and you want to know the 50th term quickly. You'd use its explicit formula, which gives you direct access to any term.
In this example, the explicit formula is \(a_n = \frac{(-1)^n \, n}{2n - 1}\). This formula is crafted by observing patterns:

• The numerators \((-1), 2, (-3), 4, \ldots\) alternate in sign, defined as \((-1)^n n\).
• The denominators \(1, 3, 5, 7, \ldots\) are consecutive odd numbers, captured by \(2n - 1\).

Using this rule, you can calculate any term by simply substituting \(n\) with the desired position number. This saves you time and effort, providing a powerful way to analyze the behavior of sequences over an endless horizon.

Sequence Convergence

Convergence in sequences is an interesting property. It tells us whether a sequence "settles down" to a single value as it stretches to infinity. We say a sequence converges if it approaches a specific finite number. Otherwise, it diverges.
In our example, understanding whether \(a_n = \frac{(-1)^n \, n}{2n - 1}\) converges or not involves looking at how the sequence behaves as \(n\) increases:

• The fraction \(\frac{n}{2n-1}\) is crucial. As \(n\) gets very large, the denominator and numerator grow, but the \(2n\) in the denominator keeps it larger than the numerator.
• Practically, the sequence becomes \(\frac{1}{2}\) as the impact of the \(-1\) becomes negligible.

This behavior indicates convergence towards a single value, allowing us to say the sequence indeed converges. Recognizing convergence is like finding stability in an ever-growing sequence, a valuable insight in mathematics.

Limits of Sequences

When discussing the limit of sequences, we're essentially looking at the long-term behavior of the sequence as \(n\) approaches infinity. The limit provides a target value that the sequence gets closer and closer to.
For the sequence \(a_n = \frac{(-1)^n \, n}{2n - 1}\), its limit can be calculated as \(n\) goes to infinity:

• The gift of limits lies in simplifying expressions by zeroing in on dominant terms. Here, \((2n - 1)\) in the denominator barely grows more than \(n\), making \(\frac{n}{2n-1} \approx \frac{1}{2}\).
• This tells us that as \(n\) increases without bound, the sequence approaches \(\frac{1}{2}\).
• Thus, the limit of the sequence \( \lim_{n \to \infty} a_n \) is \(\frac{1}{2}\), offering a precise number that encapsulates the sequence's endless journey.

Limits elegantly capture the final destination of a sequence, delivering clarity into what might otherwise seem like chaotic changes.