Problem 58
Question
Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=(n+4)^{1 /(n+4)} $$
Step-by-Step Solution
Verified Answer
The sequence \( a_n = (n+4)^{1/(n+4)} \) converges to 1.
1Step 1: Analyze the Sequence Definition
The sequence given is \( a_n = (n+4)^{1/(n+4)} \). We need to determine whether this sequence converges or diverges as \( n \) approaches infinity.
2Step 2: Calculate the Limit as n Approaches Infinity
To find the limit of \( a_n \) as \( n \to \infty \), consider \( L = \lim_{n \to \infty} (n+4)^{1/(n+4)} \). We take the natural logarithm to simplify: \( \ln L = \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} \).
3Step 3: Apply L'Hôpital's Rule
Since \( \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} \) is an indeterminate form \( \frac{\infty}{\infty} \), apply L'Hôpital's Rule by differentiating the numerator and the denominator: \( \lim_{n\to\infty} \frac{\ln(n+4)}{n+4} = \lim_{n\to\infty} \frac{\frac{d}{dn}[\ln(n+4)]}{\frac{d}{dn}[n+4]} = \lim_{n\to\infty} \frac{1/(n+4)}{1} = \lim_{n\to\infty} \frac{1}{n+4} = 0 \).
4Step 4: Conclude the Limit of the Original Sequence
Since \( \ln L = 0 \), it follows that \( L = e^0 = 1 \). Therefore, the limit of the sequence \( a_n \) is 1, and the sequence converges.
Key Concepts
Limit of a SequenceL'Hôpital's RuleNatural Logarithm
Limit of a Sequence
The concept of a limit for a sequence is crucial in understanding whether the values of the sequence get closer to a certain number as the sequence progresses. A sequence, denoted by \( a_n \), will converge if its terms get closer to a specific value \( L \), called the limit, as \( n \) - the index - goes to infinity.
To find the limit of a sequence, you analyze its behavior for large values of \( n \). Essentially, you want to find if there exists a real number \( L \) such that no matter how tiny \( \varepsilon > 0 \) you pick, there is a number \( N \) such that for all \( n > N \), the difference between the sequence value \( a_n \) and its limit \( L \) is less than \( \varepsilon \). In mathematical terms, this means \( |a_n - L| < \varepsilon \).
This property ensures that the sequence stabilizes around a number as \( n \) grows. If the sequence does not satisfy these criteria, it diverges, meaning it does not settle on a single real number.
To find the limit of a sequence, you analyze its behavior for large values of \( n \). Essentially, you want to find if there exists a real number \( L \) such that no matter how tiny \( \varepsilon > 0 \) you pick, there is a number \( N \) such that for all \( n > N \), the difference between the sequence value \( a_n \) and its limit \( L \) is less than \( \varepsilon \). In mathematical terms, this means \( |a_n - L| < \varepsilon \).
This property ensures that the sequence stabilizes around a number as \( n \) grows. If the sequence does not satisfy these criteria, it diverges, meaning it does not settle on a single real number.
L'Hôpital's Rule
L'Hôpital's Rule is a fantastic tool for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms, common in sequence limits, can't be solved through simple substitution.
The rule states that for functions \( f(x) \) and \( g(x) \), the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) can be evaluated as \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided this latter limit exists. The derivatives of the numerator and denominator are used, keeping the limit point \( c \) the same.
In our exercise, we encounter an indeterminate form \( \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} \). By applying L'Hôpital's Rule, we differentiate \( \ln(n + 4) \), yielding \( \frac{1}{n+4} \), and \( n + 4 \), yielding \( 1 \). We then find the limit: \( \lim_{n \to \infty} \frac{1}{n+4} = 0 \). Thus, L'Hôpital's Rule helps us conclude that the limit of the natural logarithm part is zero, simplifying our main limit problem.
The rule states that for functions \( f(x) \) and \( g(x) \), the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) can be evaluated as \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided this latter limit exists. The derivatives of the numerator and denominator are used, keeping the limit point \( c \) the same.
In our exercise, we encounter an indeterminate form \( \lim_{n \to \infty} \frac{\ln(n+4)}{n+4} \). By applying L'Hôpital's Rule, we differentiate \( \ln(n + 4) \), yielding \( \frac{1}{n+4} \), and \( n + 4 \), yielding \( 1 \). We then find the limit: \( \lim_{n \to \infty} \frac{1}{n+4} = 0 \). Thus, L'Hôpital's Rule helps us conclude that the limit of the natural logarithm part is zero, simplifying our main limit problem.
Natural Logarithm
Natural logarithms, notated as \( \ln(x) \), describe how many times you should multiply the number \( e \) (approximately \( 2.71828 \)) to get \( x \). The natural logarithm is crucial for simplifying expression related to growth and decay which are prevalent in many mathematical and physical contexts.
In dealing with limits, natural logarithms are often used to translate potentially complex expressions into a more manageable form. Specifically, when finding limits that involve expressions raised to a power, taking the natural logarithm can help break down the form into something you can solve more easily.
In the exercise discussed, we introduced \( \ln \) to the sequence \( (n+4)^{1/(n+4)} \) to transform it into \( \frac{\ln(n+4)}{n+4} \). This transformation allowed us to utilize L'Hôpital's Rule to solve the limit problem. Consequently, understanding natural logarithms and their properties is key to solving such sequence problems efficiently.
In dealing with limits, natural logarithms are often used to translate potentially complex expressions into a more manageable form. Specifically, when finding limits that involve expressions raised to a power, taking the natural logarithm can help break down the form into something you can solve more easily.
In the exercise discussed, we introduced \( \ln \) to the sequence \( (n+4)^{1/(n+4)} \) to transform it into \( \frac{\ln(n+4)}{n+4} \). This transformation allowed us to utilize L'Hôpital's Rule to solve the limit problem. Consequently, understanding natural logarithms and their properties is key to solving such sequence problems efficiently.
Other exercises in this chapter
Problem 58
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