Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find an explicit formula a \(a_{n}=\) ____ for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ \frac{1}{2^{2}}, \frac{2}{2^{3}}, \frac{3}{2^{4}}, \frac{4}{2^{5}}, \ldots $$

Step-by-Step Solution

Answer

The sequence's formula: \(a_n = \frac{n}{2^{n+1}}\); it converges to 0.

Step 1: Identify the General Term

Start by recognizing the pattern of the sequence and identifying the general term formula. Observing the sequence \( \frac{1}{2^{2}}, \frac{2}{2^{3}}, \frac{3}{2^{4}}, \frac{4}{2^{5}}, \ldots \), we can see a pattern where the numerator seems to be increasing by 1 and the denominator is a power of 2. Specifically, at term \(a_n\), the numerator is \(n\) and the denominator is \(2^{n+1}\). Therefore, the general formula for the sequence is:\[a_n = \frac{n}{2^{n+1}}.\]

Step 2: Determine Convergence or Divergence

To determine the convergence or divergence of the sequence, analyze the limit of \(a_n\) as \(n\) approaches infinity. The term \(\frac{n}{2^{n+1}}\) consists of a linear numerator \(n\) and an exponential denominator \(2^{n+1}\), suggesting that the exponential growth in the denominator will dominate the linear growth of the numerator. Thus, we consider the limit:\[\lim_{n \to \infty} \frac{n}{2^{n+1}}.\] Applying L'Hopital's Rule or simply recognizing the relative growth rates, this limit evaluates to 0.

Step 3: Conclusion

Since \(\lim_{n \to \infty} \frac{n}{2^{n+1}} = 0\), the sequence converges to 0. The explicit formula for the sequence is \(a_n = \frac{n}{2^{n+1}}\), and the limit of the sequence as \(n\) approaches infinity is 0.

Key Concepts

Explicit Formula

An explicit formula is a powerful tool that provides directly the nth term of a sequence without the need to rely on previous terms. In our example sequence, the pattern is revealed through observation: the numerator increases by 1 with each term, while the denominator follows the rule of powers of 2. By capturing these patterns, we can derive the explicit formula for the sequence: \ \[a_n = \frac{n}{2^{n+1}}.\]\ This formula allows us to plug in any value of \( n \) to find the term at that position, making it highly useful for calculating terms efficiently and exploring the sequence's properties.

Limit of a Sequence

The limit of a sequence, denoted as \( \lim_{n \to \infty} a_n \), is essential in understanding the behavior of a sequence as it progresses towards infinity. For our sequence, \( a_n = \frac{n}{2^{n+1}} \), we look at what value \( a_n \) approaches as \( n \) becomes very large. In our problem, the limit evaluates to: \ \[ \lim_{n \to \infty} \frac{n}{2^{n+1}} \]\ By comparing terms as \( n \) increases, we see that the numerator grows linearly, whereas the denominator grows exponentially. This understanding hints at a diminishing result towards zero. The analysis of this limit tells us that the sequence converges, and the value it approaches is indeed 0.

Exponential Growth

Exponential growth refers to a process where quantities increase at a consistent growth rate, characterized by variable exponents. In the sequence \( a_n = \frac{n}{2^{n+1}} \), the denominator \( 2^{n+1} \) showcases exponential growth.\ As \( n \) increases, \( 2^{n+1} \) increases dramatically compared to the linear growth of \( n \). This rapid increase in the denominator causes the overall value of the fraction to decrease substantially, illustrating a classic case of exponential growth where larger exponents lead to much larger outputs, overshadowing the numerator.

L'Hopital's Rule

L'Hopital's Rule is a valuable calculus tool used to evaluate limits of indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). To solve the limit \( \lim_{n \to \infty} \frac{n}{2^{n+1}} \), recognizing an indeterminate form can guide us to apply L'Hopital's Rule.\ This rule states that if the limit of a function \( \frac{f(n)}{g(n)} \) results in an indeterminate form, we can differentiate the numerator and the denominator separately and reevaluate the limit: \ \[ \lim_{n \to \infty} \frac{n}{2^{n+1}} = \lim_{n \to \infty} \frac{1}{(\ln 2)2^{n}} = 0 \]\ After differentiation, it's clear that the denominator's exponential nature dominates, leading the sequence towards zero. Applying L'Hopital's Rule gives clarity and accuracy in finding the limit, confirming the sequence's convergence.