Problem 54 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\left(1-\frac{1}{n}\right)^{n} $$

Step-by-Step Solution

Answer

The sequence converges to \( \frac{1}{e} \).

Step 1: Understanding the Sequence

The given sequence is \( a_n = \left(1 - \frac{1}{n}\right)^n \). We need to determine if this sequence converges and, if so, find its limit. As \( n \) approaches infinity, this is a classic example involving the number \( e \).

Step 2: Recognizing the Limit Expression

The expression \( \left(1 - \frac{1}{n}\right)^n \) is related to the well-known expression for \( e^{-1} \). We know that \( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \), so \( \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = e^{-1} \).

Step 3: Applying the Limit

As \( n \to \infty \), the sequence \( \left(1 - \frac{1}{n}\right)^n \) converges. We recognize this limits to \( e^{-1} \), as derived in calculus through the use of the limit definition of \( e \).

Step 4: Conclusion

Therefore, the sequence converges to \( e^{-1} \). We conclude that the sequence is convergent with the limit equal to \( \frac{1}{e} \).

Key Concepts

Exponential Limit

An exponential limit is a type of limit where exponential expressions are involved. In our exercise, we are considering the limit of a sequence expressed as \( \left(1 - \frac{1}{n}\right)^n \). As \( n \) becomes very large, understanding how this sequence behaves can help us determine if it converges to a particular value. In calculus, it is established that \( \left(1 + \frac{1}{n}\right)^n \rightarrow e \) as \( n \rightarrow \infty \), where \( e \approx 2.71828 \) is the base of the natural logarithm. A similar structure can be found in the expression \( \left(1 - \frac{1}{n}\right)^n \), which approximates \( e^{-1} \) when \( n \) becomes infinitely large. Here, the negative sign indicates that we have an inverse.This result is critical because it shows the power of limits in managing exponential expressions. We can predict and compute the behavior of sequences involving very tiny numbers multiplied multiple times, which might otherwise seem complex.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence or multi-dimensional array of values. Each term is defined as a function of one or more of its previous terms. For instance, in some problems, you'll find sequences where each term is constructed based on the previous term in a predictable manner. This occurs frequently in algorithms and mathematical modeling.In contrast, the expression \( a_n = \left(1 - \frac{1}{n}\right)^n \) doesn't explicitly use a recurrence relation. Instead, it uses a direct formula for each term \( a_n \). However, understanding recurrence relations is valuable as they form the backbone for computational approaches used to solve sequential or progression problems. Recognizing how such sequences unfold, even without explicit recurrence, allows us better algebraic manipulations in solving equations that are built on iterative processes.

Limit of a Sequence

The limit of a sequence is the value that the terms of the sequence "approach" as the index of the terms goes to infinity. If a sequence has a limit, it is considered convergent; if it doesn't have a limit, it is divergent.In the exercise, we evaluated the sequence \( a_n = \left(1 - \frac{1}{n}\right)^n \). We discovered that as \( n \rightarrow \infty \), the sequence approaches a definite value. Recognizing it converges to \( e^{-1} \) is crucial because it offers insight into the ultimate behavior of these terms.When dealing with limits, certain techniques can be used to analyze them, such as:

• Recognizing special forms that approximate known constants like \( e \).
• Applying the squeeze theorem, if necessary, to "squeeze" the sequence between two other sequences that have the same limit.

These strategies allow us to ensure that the sequence indeed behaves consistently as it extends towards infinity, solidifying our understanding of convergent series.