Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find an explicit formula a \(a_{n}=\) ____ for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ 2,1, \frac{2^{3}}{3^{2}}, \frac{2^{4}}{4^{2}}, \frac{2^{5}}{5^{2}}, \ldots $$

Step-by-Step Solution

Answer

The sequence diverges; it has no finite limit.

Step 1: Analyze the given sequence

Examine the sequence provided: \(2, 1, \frac{2^3}{3^2}, \frac{2^4}{4^2}, \frac{2^5}{5^2}, \ldots\). Notice that each term looks in the form of \(\frac{2^n}{n^2}\), starting from the second term.

Step 2: Formulate the explicit formula

Based on the sequence pattern, the terms can be expressed as \(a_n = \frac{2^n}{n^2}\) for \(n \geq 1\). This formula accurately matches each term in the sequence.

Step 3: Determine convergence or divergence

To determine if the sequence converges or diverges, analyze the behavior of the formula \(a_n = \frac{2^n}{n^2}\) as \(n\) approaches infinity. Compare the growth rates of the numerator and the denominator. The term \(2^n\) grows exponentially which is faster than the polynomial growth of \(n^2\), indicating that \(a_n\) diverges to infinity.

Step 4: Conclude the behavior of the sequence

Given the faster exponential growth of \(2^n\) compared to the polynomial \(n^2\), the sequence diverges. Therefore, there is no limit as \(n\) approaches infinity.

Key Concepts

Convergence and Divergence

When analyzing sequences, one major step is to determine if they converge or diverge. A sequence converges if its terms tend to approach a single, specific value as the sequence progresses. In contrast, the sequence diverges when the terms continue growing indefinitely, or don't settle towards a singular value.
In this exercise, the sequence is expressed through the formula \(a_n = \frac{2^n}{n^2}\). Here, to determine convergence or divergence, we observe how the sequence behaves as \(n\) becomes very large.
The term \(2^n\) grows exponentially, meaning it multiplies by two with each step, while the denominator \(n^2\) grows as a polynomial, squaring the number \(n\) each time. Since exponential functions grow much faster than polynomial functions, this sequence diverges, as it doesn't approach a finite limit. It effectively means the sequence maintains growth without bound.

Explicit Formula

Writing an explicit formula is like creating a blueprint for a sequence. It provides a clear, direct method to calculate any term of the sequence without having to sum terms or follow iterative processes.
For the sequence \(2, 1, \frac{2^3}{3^2}, \frac{2^4}{4^2}, \frac{2^5}{5^2}, \ldots\), each term can be described systematically using the formula \(a_n = \frac{2^n}{n^2}\), applied for \(n \geq 1\).
This formula stems from observing the pattern where the numerator is \(2\) raised to the power \(n\) and the denominator is the square of the term's position, \(n\). Having an explicit formula helps not only in predicting the behavior of sequences but also in evaluating terms swiftly, without the need for recursive calculations.

Exponential Growth

Exponential growth in sequences is characterized by the rapid increase in the value of terms. Whenever we see an expression like \(2^n\), it signifies exponential growth, as each term is double the previous one. In general, exponential functions look like \(b^n\), with \(b\) being a base bigger than 1.
In our sequence, the \(2^n\) in the numerator is a prime example of exponential growth. This kind of growth means that as \(n\) becomes large, \(2^n\) expands incredibly fast, much quicker than any linear or polynomial growth, such as \(n^2\).
This dramatic rate of increase is why sequences involving exponential terms often diverge unless a strong limiting factor appears to counterbalance this growth. In our specific sequence analysis, without a controlling factor, the exponential growth leads to a divergence, meaning the sequence doesn't settle at a particular value.