Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{n^{100}}{e^{n}}\)

Step-by-Step Solution

Answer

First five terms: approximately 0.368, large numbers. Sequence converges, limit is 0.

Step 1: Calculate the First Five Terms

To find the first five terms of the sequence \( \{a_n\} = \left( \frac{n^{100}}{e^n} \right) \), we substitute \( n \) with 1 to 5. \[a_1 = \frac{1^{100}}{e^{1}} = \frac{1}{e} \approx 0.368a_2 = \frac{2^{100}}{e^{2}} = \frac{1267650600228229401496703205376}{7.389} \approx 1.715 \times 10^{25}a_3 = \frac{3^{100}}{e^{3}} \approx 5.151 \times 10^{46}a_4 = \frac{4^{100}}{e^{4}} \approx 3.000 \times 10^{64} a_5 = \frac{5^{100}}{e^{5}} \approx 2.403 \times 10^{79}\]

Step 2: Determine Convergence or Divergence

A sequence \( a_n = \frac{n^{100}}{e^n} \) can be analyzed for convergence by comparing the growth rates of the numerator and denominator. The denominator, \( e^n \), grows exponentially, while the numerator, \( n^{100} \), grows polynomially. Since the exponential growth of \( e^n \) eventually outpaces the polynomial growth of \( n^{100} \), the sequence \( a_n \to 0 \) as \( n \to \infty \). Therefore, the sequence converges.

Step 3: Calculate the Limit

Since \( a_n = \frac{n^{100}}{e^n} \) converges, we find \( \lim_{n \to \infty} a_n \). Comparing polynomial and exponential growth, we apply L'Hopital's Rule or recognize that the exponential term dominates:\[\lim_{n \to \infty} \frac{n^{100}}{e^n} = 0\]Thus, the limit of the sequence is 0.

Key Concepts

Limit of a Sequence

The concept of the limit of a sequence is a fundamental one in calculus and analysis. It is essentially about understanding the behavior of a sequence as its terms progress towards infinity. In simpler terms, we want to know what value a sequence approaches as the index, usually denoted by \( n \), becomes very large.
When we say a sequence \( \{a_n\} \) converges to a limit \( L \), this means that as \( n \) increases, the terms \( a_n \) get arbitrarily close to \( L \). If such a \( L \) exists, we write \( \lim_{n \to \infty} a_n = L \).
Importantly, if no such value exists, the sequence is said to diverge. In the given exercise where the sequence is \( a_n = \frac{n^{100}}{e^n} \), we find that \( a_n \) tends to 0 as \( n \) becomes very large, indicating that the sequence converges to 0. This conclusion helps us understand the overall behavior indicating a "settling" towards 0.

Polynomial vs Exponential Growth

Understanding the difference between polynomial and exponential growth is key in assessing limits of sequences like \( \{a_n\} = \frac{n^{100}}{e^n} \). Let's break down how these growth rates differ:
- **Polynomial growth** refers to functions like \( n^k \) where \( k \) is a constant. The growth is significant, but relatively slower as \( n \) increases, especially in comparison to exponential functions.
- **Exponential growth**, such as \( e^n \), means that the quantity grows by a fixed multiplier at each step and thus accelerates very rapidly as \( n \) increases.
The problem at hand shows \( n^{100} \) as polynomial and \( e^n \) as exponential. Given enough time (i.e., as \( n \to \infty \)), \( e^n \) will grow much faster than \( n^{100} \), causing the ratio \( \frac{n^{100}}{e^n} \) to approach 0. Hence, we conclude that the sequence converges to 0 due to this dominance of exponential growth over polynomial growth.

L'Hopital's Rule

L'Hopital's Rule is a handy tool in calculus used to find the limits of indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It states that if these forms are encountered, one can compute the derivative of the numerator and the derivative of the denominator, then find their limit instead, under certain conditions.
In the exercise \( a_n = \frac{n^{100}}{e^n} \), if directly evaluating the limit as \( n \to \infty \) gives an indeterminate form like \( \frac{\infty}{\infty} \), we can apply L'Hopital's Rule multiple times until a determinate form is achieved. Ultimately, in cases where exponential growth is present, L'Hopital's Rule (or recognizing the nature of growth rates) will reveal that the polynomial term vanishes in face of exponentially outpacing growth, justifying \( \lim_{n \to \infty} a_n = 0 \). Thus, L'Hopital's Rule simplifies solving complex limits and confirms convergence in this scenario.