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. \(a_{n}=\frac{(2 n+1)^{2 n}}{\left(3 n^{2}+1\right)^{n}}\)

Short answer

The limit does not exist as the sequence diverges.

Step-by-step solution

Analyze the Sequence

The given sequence is \(a_n = \frac{(2n+1)^{2n}}{(3n^2+1)^n}\). To apply the divergence test, we need to determine whether \(\lim_{n \to \infty} a_n\) exists or not.

Simplify the Sequence

Rewrite the sequence for large \(n\) terms: \(a_n = \left(\frac{(2n+1)^{2}}{3n^2+1}\right)^n\). This can be further expressed as \(\left(\frac{4n^2 + 4n + 1}{3n^2 + 1}\right)^n\).

Estimate Leading Behavior of Numerator and Denominator

For large \(n\), the leading terms in both the numerator and denominator dominate, thus \(\frac{4n^2 + 4n + 1}{3n^2 + 1} \approx \frac{4n^2}{3n^2} = \frac{4}{3}\).

Calculate the Limit Manipulating the Exponent

The expression simplifies to \(\left(\frac{4}{3}\right)^n\). Since \(\frac{4}{3} > 1\), \(\left(\frac{4}{3}\right)^n\) grows without bounds as \(n \to \infty\).

Conclusion about the Limit

As \(n\) approaches infinity, \(a_n\) clearly diverges because the base \(\frac{4}{3}\) is greater than 1 and the exponent is \(n\). Thus, \(\lim_{n \to \infty} a_n\) does not exist.

Key concepts

Limit of a Sequence

A sequence is like a list of numbers where each number is determined by its position in the list. Sometimes, when you go far enough in the sequence, the numbers start getting really close to a specific value. This specific value is what we call the limit of the sequence. In mathematical terms, for a sequence \( \{a_n\} \), the limit is the value that \( a_n \) approaches as \( n \) becomes very large.For the sequence given in the exercise, \( a_n = \frac{(2n+1)^{2n}}{(3n^2+1)^n} \), we want to find out if there's a limit as \( n \) goes to infinity. If the sequence approaches a finite number, then that is its limit. However, if the numbers get larger without bound, the sequence diverges, meaning it doesn't have a limit.In the solved problem, the expression simplifies to \( \left(\frac{4}{3}\right)^n \). Since the base is greater than 1, the sequence grows without limit as \( n \) increases. Thus, the limit does not exist.

Asymptotic Analysis

Asymptotic analysis is a way to describe the behavior of sequences or functions as the input grows very large. It's like predicting the end behavior of the sequence. In the context of this exercise, it involves simplifying the sequence expression by focusing on the most significant terms as \( n \) approaches infinity.For example, in \( a_n = \frac{(2n+1)^{2n}}{(3n^2+1)^n} \), the numerator and denominator have leading terms that dominate when \( n \) is large. By analyzing these terms, we conclude that \( \frac{4n^2 + 4n + 1}{3n^2 + 1} \approx \frac{4n^2}{3n^2} = \frac{4}{3} \).This approximation simplifies the sequence's behavior to \( \left(\frac{4}{3}\right)^n \). This method, the asymptotic analysis, helps us deduce that as \( n \to \infty \), the sequence diverges because it does not approach a finite limit.

Exponential Growth

Understanding exponential growth is key here. When a quantity increases exponentially, it grows at a rate proportional to its current value. This means, as time (or, in our case, \( n \)) increases, the quantity grows faster and faster.The expression \( \left(\frac{4}{3}\right)^n \) is an example of exponential growth. Because the base \( \frac{4}{3} \) is greater than 1, each term in the sequence is larger than the previous one. - Exponential growth is characterized by fast increase.- The larger \( n \), the more rapidly the sequence values increase.- As \( n \) goes to infinity, the values do not settle to any finite number.In practical terms, this kind of growth is why \( a_n = \left(\frac{4}{3}\right)^n \) diverges. It does not stabilize, rather it shoots up towards infinity, emphasizing the lack of a limit for the sequence.