Problem 23 - Essential Calculus Early Transcendentals

Question

\(9-32\) n Determine whether the sequence converges or diverges. If it converges, find the limit. $$\left\\{n^{2} e^{-n}\right\\}$$

Step-by-Step Solution

Answer

The sequence converges to 0 as \(n\) approaches infinity.

Step 1: Understanding the Sequence

The sequence presented is \(a_n = n^2 e^{-n}\). Our task is to determine if this sequence converges or diverges and if it converges to find the limit.

Step 2: Rewrite with Exponentials

Rewrite the term \(e^{-n}\) as \(\frac{1}{e^n}\). Thus, the sequence becomes \(a_n = \frac{n^2}{e^n}\).

Step 3: Consider the Growth Rates

Consider the growth rates of the terms: \(n^2\) grows polynomially, whereas \(e^n\) grows exponentially. Exponential growth is much faster than polynomial growth as \(n\) approaches infinity.

Step 4: Apply Limit Laws

To analyze the limit, compute \(\lim_{n \to \infty} \frac{n^2}{e^n}\).

Step 5: L'Hôpital's Rule Application

Direct substitution results in an indeterminate form \(\frac{\infty}{\infty}\). Differentiate the numerator and the denominator separately: the derivative of \(n^2\) is \(2n\), and the derivative of \(e^n\) is \(e^n\).

Step 6: Evaluate the New Limit

The limit now is \(\lim_{n \to \infty} \frac{2n}{e^n}\). Again, applying L'Hôpital's Rule, differentiate the numerator \(2n\) to get \(2\), and \(e^n\) remains \(e^n\).

Step 7: Final Limit Calculation

Calculate \(\lim_{n \to \infty} \frac{2}{e^n}\), which evaluates to zero as \(e^n\) grows much faster than the constant numerator.

Step 8: Conclusion on Convergence

Since the limit of \(a_n\) is 0, the sequence \(\{n^2 e^{-n}\}\) converges to 0.

Key Concepts

Exponential Growth

Exponential growth is a fascinating concept where quantities increase at a rate proportional to their size. The function \(e^n\) is a classic example of exponential growth. As \(n\) becomes larger, \(e^n\) increases rapidly, much faster than any polynomial function like \(n^2\). This rapid growth has practical implications in numerous fields.
In nature, you can see exponential growth in populations, where each generation breeds exponentially more individuals. In technology, data storage requirements often increase exponentially as well.

• Exponential growth functions describe many real-life phenomena, including finance, biology, and technology.
• In calculus, exponential growth often plays a vital role in determining the behavior of sequences and mathematical models.

Understanding how exponential growth compares with other types of growth is crucial when analyzing limits and sequences.

L'Hôpital's Rule

L'Hôpital's Rule is a handy tool in calculus for evaluating limits involving indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It involves taking the derivative of the numerator and denominator until a determinate form is attained.
When using L'Hôpital's Rule:

• Confirm the limit is indeed an indeterminate form.
• Differentiate the numerator and the denominator separately and then take the limit again.
• If the resultant limit remains indeterminate, apply L'Hôpital's Rule repetitively.

This rule makes complex limits manageable and provides a pathway to solving difficult sequence problems.
In our example, we used L'Hôpital's Rule twice to transform \(\lim_{n \to \infty} \frac{n^2}{e^n}\) to \(\lim_{n \to \infty} \frac{2}{e^n}\), where it evaluates to zero, confirming convergence.

Limits and Infinity

When analyzing sequences, especially those involving growth, limits and infinity are essential concepts. A limit helps us determine a value as a function "approaches" a point, often infinity. The idea of infinity in limits is abstract and deals with the behavior of sequences as variables grow unbounded.
To understand limits at infinity:

• Consider what happens to the value of a sequence as \(n\) grows larger and larger.
• If each term in the sequence gets closer to a particular number, the sequence is said to converge to that limit.
• If not, it diverges, going off toward infinity or negative infinity.

In our sequence, \(\{n^2 e^{-n}\}\), the limit as \(n\) approaches infinity was found to be 0, confirming convergence. This demonstrates that even though \(n^2\) increases, the dominance of \(e^n\) ensures the sequence approaches zero at infinity.