Question

For each of the following sequences, if the divergence test applies, either state that \(\lim _{n \rightarrow \infty} a_{n}\) does not exist or find \(\lim _{n \rightarrow \infty} a_{n} .\) If the divergence test does not apply, state why. \(a_{n}=e^{-2 / n}\)

Short answer

The divergence test applies; the series diverges.

Step-by-step solution

Understand the Divergence Test

The divergence test states that if the limit of a sequence \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum a_n \) diverges. If \( \lim_{n \to \infty} a_n = 0 \), the test is inconclusive.

Calculate the Limit of the Sequence

We are given the sequence \( e^{-2/n} \). To find the limit as \( n \to \infty \), we calculate: \[ \lim_{n \to \infty} e^{-2/n} = e^{\lim_{n \to \infty} -2/n} \].

Evaluate the Inside Limit

The expression inside the exponent is \( -2/n \). As \( n \to \infty \), \( -2/n \to 0 \). Thus, \[ \lim_{n \to \infty} -2/n = 0 \].

Apply the Limit to the Exponential Function

Substitute the result from Step 3 into the exponential function: \[ \lim_{n \to \infty} e^{-2/n} = e^{0} = 1 \].

Analyze the Result

The limit \( \lim_{n \to \infty} a_n = 1 \) is not equal to zero, thus the divergence test applies, and the conclusion is that the series diverges.

Key concepts

Sequences and Series

Understanding sequences and series is crucial when analyzing patterns in mathematics. A **sequence** is an ordered list of numbers, each of which is called a term. An important aspect of sequences is determining their behavior as they progress. In other words, what happens as you go further out into the sequence?A **series**, on the other hand, is the sum of the terms of a sequence. When we talk about convergence or divergence in this context, we're generally referring to series. A series can either converge to a specific value or fail to converge (diverge). To work with sequences and series, we commonly use tests like the Divergence Test to assess the behavior. For example, in our exercise, we analyze the sequence defined by the terms \(a_n = e^{-2/n}\). Through our calculations, we determined the sequence itself converges to a limit of 1; however, as a series, since the terms do not tend to 0, it diverges.

Limit Evaluation

Limit evaluation is a fundamental concept when dealing with sequences. It involves finding the value that the terms of a sequence approach as the index \(n\) becomes very large. In mathematical terms, we write this as \(\lim_{n \to \infty} a_n\). Evaluating this limit gives us a crucial clue about the behavior of the sequence over time.To perform limit evaluation for our sequence \(a_n = e^{-2/n}\), we examined the expression inside the exponential function, which was \(-2/n\). By observing what happens as \(n\) approaches infinity, we saw that \(-2/n\) tends to zero.This allowed us to conclude: - \(\lim_{n \to \infty} e^{-2/n} = e^{0} = 1\)Understanding how to manipulate and calculate limits is vital for anyone exploring higher mathematics, as it provides insights into the long-term behavior of sequences.

Exponential Functions

Exponential functions play a significant role in both sequences and series. These functions have the form \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828.Exponential functions are powerful because they model growth and decay processes very effectively. In our specific problem with the sequence \(a_n = e^{-2/n}\), we used properties of exponential functions to determine the sequence's limit.Here’s how they help:- Exponential functions transform even complex expressions into something more manageable.- They turn limits of fractions and negative powers into easy-to-interpret growth behaviors.When \(x\) is very small, as is the case of \(-2/n\) when \(n\) is large, the exponential function \(e^x\) approaches 1. This is why \(e^0 = 1\), simplifying limit evaluations for sequences involving exponential forms.