Problem 62
Question
Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=(n+4)^{1 /(n+4)} $$
Step-by-Step Solution
Verified Answer
The sequence converges, with a limit of 1.
1Step 1: Understand the Concept of Convergence
A sequence \( \{a_n\} \) converges if its terms approach a specific value as \( n \to \infty \). The limit \( L \) of a sequence is this specific value that \( a_n \) approaches. If no such value exists, the sequence diverges.
2Step 2: Analyze the Sequence \( a_n \)
The sequence given is \( a_n = (n+4)^{1/(n+4)} \). As \( n \to \infty \), we should determine whether the terms of this sequence approach a specific value (i.e., find \( \lim_{n \to \infty} a_n \)).
3Step 3: Apply the Limit Definition
To analyze convergence, compute \( \lim_{n \to \infty} a_n \). Rewrite \( a_n \) as \( a_n = (n+4)^{1/(n+4)} = e^{(1/(n+4)) \ln(n+4)} \). You're converting a power into an exponential form to take limits easily.
4Step 4: Simplify the Exponential Expression
Focus on the expression \( (1/(n+4)) \ln(n+4) \). As \( n \to \infty \), apply L'Hopital's Rule to \( \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} \) since it is in an indeterminate form \( \frac{\infty}{\infty} \).
5Step 5: Apply L'Hopital's Rule
Differentiate the numerator and denominator: The derivative of \( \ln(n+4) \) is \( \frac{1}{n+4} \) and the derivative of \( n+4 \) is \( 1 \). Thus, L'Hopital's Rule gives: \[ \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} = \lim_{n \to \infty} \frac{1/(n+4)}{1} = \lim_{n \to \infty} \frac{1}{n+4} = 0 \].
6Step 6: Determine the Convergence of the Sequence
Since \( \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} = 0 \), we conclude that \( \lim_{n \to \infty} e^{(1/(n+4)) \ln(n+4)} = e^0 = 1 \).
7Step 7: Conclusion on the Sequence
The sequence \( a_n = (n+4)^{1/(n+4)} \) converges, and the limit of the sequence is 1.
Key Concepts
Limit of a SequenceL'Hôpital's RuleIndeterminate FormsExponential Functions
Limit of a Sequence
The limit of a sequence is an essential concept in calculus. It helps us determine whether a sequence approaches a specific value as the number of terms goes to infinity. Essentially, when we talk about the limit of a sequence \( \{a_n\} \), we are interested in finding a number \( L \) such that the terms \( a_n \) get arbitrarily close to \( L \) as \( n \) increases. If such a number exists, the sequence is said to converge to \( L \). If no such number exists, the sequence diverges.
In our exercise, the sequence is given by \( a_n = (n+4)^{1/(n+4)} \). To find its limit, we need to determine the behavior of \( a_n \) as \( n \to \infty \). By rewriting \( a_n \) as an exponential function, we can explore its limit more effectively, leading to convergence at a specific point.
In our exercise, the sequence is given by \( a_n = (n+4)^{1/(n+4)} \). To find its limit, we need to determine the behavior of \( a_n \) as \( n \to \infty \). By rewriting \( a_n \) as an exponential function, we can explore its limit more effectively, leading to convergence at a specific point.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus, particularly when dealing with limits that result in indeterminate forms. This rule can be used to evaluate limits of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) by taking the derivative of the numerator and the denominator separately.
In our solution, we encountered an expression that needed simplification: \( \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} \). This expression falls into an indeterminate form \( \frac{\infty}{\infty} \), allowing us to apply L'Hôpital's Rule. By differentiating the numerator and denominator, we find:
In our solution, we encountered an expression that needed simplification: \( \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} \). This expression falls into an indeterminate form \( \frac{\infty}{\infty} \), allowing us to apply L'Hôpital's Rule. By differentiating the numerator and denominator, we find:
- Derivative of \( \ln(n+4) \) is \( \frac{1}{n+4} \).
- Derivative of \( n+4 \) is \( 1 \).
Indeterminate Forms
Indeterminate forms are expressions that arise during the evaluation of limits, where the expression does not provide enough information to determine the limit immediately. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and others like \( 0\cdot\infty \) and \( \infty - \infty \).
In our problem, when we first evaluate the limit of \( \frac{\ln(n+4)}{n+4} \), it results in an indeterminate form \( \frac{\infty}{\infty} \). This is because both \( \ln(n+4) \) and \( n+4 \) grow indefinitely as \( n \) increases. Addressing indeterminate forms often requires techniques such as simplification, rewriting expressions, or applying L'Hôpital's Rule which enables us to find the limit accurately.
Understanding how to recognize and resolve indeterminate forms is crucial in calculus, as they often show up in problems involving limits.
In our problem, when we first evaluate the limit of \( \frac{\ln(n+4)}{n+4} \), it results in an indeterminate form \( \frac{\infty}{\infty} \). This is because both \( \ln(n+4) \) and \( n+4 \) grow indefinitely as \( n \) increases. Addressing indeterminate forms often requires techniques such as simplification, rewriting expressions, or applying L'Hôpital's Rule which enables us to find the limit accurately.
Understanding how to recognize and resolve indeterminate forms is crucial in calculus, as they often show up in problems involving limits.
Exponential Functions
Exponential functions are mathematical expressions involving a constant base raised to a variable exponent. They are fundamental in many areas of calculus due to their unique properties related to growth and decay.
In solving the sequence \( a_n = (n+4)^{1/(n+4)} \), we transformed the given power into an exponential function. Specifically, we wrote \( a_n \) as \( e^{(1/(n+4)) \ln(n+4)} \). This transformation allows us to take advantage of the natural logarithm and exponential properties when evaluating limits.
In solving the sequence \( a_n = (n+4)^{1/(n+4)} \), we transformed the given power into an exponential function. Specifically, we wrote \( a_n \) as \( e^{(1/(n+4)) \ln(n+4)} \). This transformation allows us to take advantage of the natural logarithm and exponential properties when evaluating limits.
- The expression \( e^{(1/(n+4)) \ln(n+4)} \) simplifies the process, making it easier to see that as \( n \to \infty \), the inner part \( (1/(n+4)) \ln(n+4) \) approaches 0.
- Thus, \( e^0 = 1 \), showing that the sequence converges to 1.
Other exercises in this chapter
Problem 62
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