Chapter 9: Problem 94
Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises converge, and which diverge? Find the limit of each convergent sequence. $$a_{n}=\sqrt[n]{n^{2}+n}$$
Short Answer
Expert verified
The sequence \( \{a_n\} \) converges to 1.
Step by step solution
01
Understand the Sequence
The given sequence is \( a_n = \sqrt[n]{n^2 + n} \). This represents the \( n \)-th root of the expression \( n^2 + n \). To determine whether this sequence converges or diverges, we will analyze the behavior as \( n \) approaches infinity.
02
Simplify the Expression
We rewrite \( n^2 + n \) as \( n^2(1 + \frac{1}{n}) \). Thus, \( a_n = \sqrt[n]{n^2(1 + \frac{1}{n})} = \sqrt[n]{n^2} \cdot \sqrt[n]{1 + \frac{1}{n}} \). This simplifies to \( n^{\frac{2}{n}} \cdot (1 + \frac{1}{n})^{\frac{1}{n}} \).
03
Analyze Each Factor
We will analyze each factor separately as \( n \to \infty \). First, consider \( n^{\frac{2}{n}} \). This can be rewritten as \( (e^{\ln n})^{\frac{2}{n}} = e^{\frac{2 \ln n}{n}} \). The exponent \( \frac{2 \ln n}{n} \) tends to 0 because \( \ln n \to \infty \) slower than \( n \to \infty \). Hence, \( n^{\frac{2}{n}} \to e^0 = 1 \).
04
Analyze the Second Factor
Now consider \( (1 + \frac{1}{n})^{\frac{1}{n}} \). Using the binomial approximation or L'Hôpital's Rule, we find that \( \ln(1 + \frac{1}{n}) \approx \frac{1}{n} \) for large \( n \). Therefore, \( \frac{1}{n} \cdot \ln(1 + \frac{1}{n}) \to 0 \) and consequently, \( (1 + \frac{1}{n})^{\frac{1}{n}} \to e^0 = 1 \).
05
Determine Overall Behavior
Both factors \( n^{\frac{2}{n}} \) and \( (1 + \frac{1}{n})^{\frac{1}{n}} \) tend to 1 as \( n \to \infty \). Therefore, the whole product \( n^{\frac{2}{n}} \cdot (1 + \frac{1}{n})^{\frac{1}{n}} \to 1 \). So, the sequence \( a_n \to 1 \) as \( n \to \infty \).
06
Conclusion on Convergence
Since \( a_n \to 1 \) as \( n \to \infty \), the sequence \( \{a_n\} \) converges. The limit of the sequence \( \{a_n\} \) is 1.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Limits and Convergence
A sequence is said to be convergent if its terms approach a specific value, called the limit, as the sequence progresses. In mathematical terms, a sequence \( \{a_n\} \) converges to a limit \( L \) if for every positive number \( \epsilon \), there exists a positive integer \( N \) such that for all \( n > N \), \( |a_n - L| < \epsilon \). This essentially means that past a certain point, the terms of the sequence get arbitrarily close to \( L \).
For the sequence \( a_n = \sqrt[n]{n^2 + n} \), we've identified that it converges. By simplifying the expression and examining the limiting behavior of its components, we found that \( a_n \) approaches 1 as \( n \to \infty \). This conclusion leverages the fundamental property of limits and the behavior of sequences, confirming that \( a_n \) settles at 1 for large \( n \). Thus, understanding limits gives us a profound insight into whether a sequence will converge or diverge.
For the sequence \( a_n = \sqrt[n]{n^2 + n} \), we've identified that it converges. By simplifying the expression and examining the limiting behavior of its components, we found that \( a_n \) approaches 1 as \( n \to \infty \). This conclusion leverages the fundamental property of limits and the behavior of sequences, confirming that \( a_n \) settles at 1 for large \( n \). Thus, understanding limits gives us a profound insight into whether a sequence will converge or diverge.
Infinity Behavior and Its Implications
When analyzing sequences, especially those involving terms like \( n^2 + n \), it is crucial to explore their behavior as \( n \) approaches infinity. Does the sequence grow without bound, approach a specific value, or vary erratically? For the sequence \( a_n = \sqrt[n]{n^2 + n} \), examining how each part behaves as \( n \) grows is key.
Upon simplifying terms in the sequence, notice that while \( n^2 + n \) itself grows without bound as \( n \to \infty \), the way it is modified by the \( n \)-th root significantly affects the result. Here, due to the properties of exponents and the logarithmic function, the components like \( n^{2/n} \) and \( (1 + 1/n)^{1/n} \) both tend to approach 1 as \( n \) becomes large. This shows how seemingly divergent expressions under specific operations can converge, highlighting a fascinating aspect of infinity behavior.
Upon simplifying terms in the sequence, notice that while \( n^2 + n \) itself grows without bound as \( n \to \infty \), the way it is modified by the \( n \)-th root significantly affects the result. Here, due to the properties of exponents and the logarithmic function, the components like \( n^{2/n} \) and \( (1 + 1/n)^{1/n} \) both tend to approach 1 as \( n \) becomes large. This shows how seemingly divergent expressions under specific operations can converge, highlighting a fascinating aspect of infinity behavior.
Exploration of the n-th Root
The \( n \)-th root is a fundamental concept in mathematics that deeply influences how sequences behave. When we talk about the \( n \)-th root, denoted as \( \sqrt[n]{x} \), we are essentially seeking a number that, when raised to the power of \( n \), yields \( x \).
In sequences like \( a_n = \sqrt[n]{n^2 + n} \), understanding the impact of the \( n \)-th root is essential. As \( n \) increases, the \( n \)-th root tends to "neutralize" the effect of polynomial growth in its argument, causing factors like \( n^{2/n} \) and \( (1 + 1/n)^{1/n} \) to converge towards 1, even if \( n^2 + n \) itself grows indefinitely. This characteristic of the \( n \)-th root provides a balancing effect, ensuring that the sequence doesn’t diverge but instead stabilizes to a finite limit.
In sequences like \( a_n = \sqrt[n]{n^2 + n} \), understanding the impact of the \( n \)-th root is essential. As \( n \) increases, the \( n \)-th root tends to "neutralize" the effect of polynomial growth in its argument, causing factors like \( n^{2/n} \) and \( (1 + 1/n)^{1/n} \) to converge towards 1, even if \( n^2 + n \) itself grows indefinitely. This characteristic of the \( n \)-th root provides a balancing effect, ensuring that the sequence doesn’t diverge but instead stabilizes to a finite limit.