Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(\frac{1}{2^{2}}, \frac{2}{2^{3}}, \frac{3}{2^{4}}, \frac{4}{2^{5}}, \ldots\)

Step-by-Step Solution

Answer

The sequence converges with the explicit formula \(a_n = \frac{n}{2^{n+1}}\), and the limit is 0.

Step 1: Identify the General Term of the Sequence

To find the general term, observe the pattern of numerators and denominators in the sequence. The sequence given is \(\frac{1}{2^2}, \frac{2}{2^3}, \frac{3}{2^4}, \frac{4}{2^5}, \ldots\). Notice that the numerator is \(n\), starting from 1, and the denominator is \(2^{n+1}\). Therefore, the general term of the sequence is \(a_n = \frac{n}{2^{n+1}}\).

Step 2: Determine Convergence or Divergence

To determine whether the sequence converges or diverges, find \(\lim_{n \to \infty} a_n\). The general term is \(a_n = \frac{n}{2^{n+1}}\). As \(n\) increases, the denominator \(2^{n+1}\) grows exponentially while the numerator \(n\) grows linearly.

Step 3: Apply Limits to Determine Convergence

Evaluate \(\lim_{n \to \infty} \frac{n}{2^{n+1}}\). Since exponential growth in the denominator outweighs the linear growth in the numerator, the limit approaches 0. Therefore, the sequence converges.

Key Concepts

Explicit Formula

An explicit formula in the context of sequences is a way to define each term of a sequence with a direct formula. This means you don't have to calculate all previous terms to find the next one; instead, you just plug the desired term's position into the formula.
This approach is highly useful because it simplifies calculations and saves time, especially for sequences with many terms.
In the original problem, the sequence given is \[ \frac{1}{2^2}, \frac{2}{2^3}, \frac{3}{2^4}, \frac{4}{2^5}, \ldots \] Observing the pattern, we see that the numerator is simply the position of the term, denoted as \(n\), and the denominator is \(2^{n+1}\). Thus, the explicit formula for this sequence is \[ a_n = \frac{n}{2^{n+1}} \] With this formula, you can find any term in the sequence without having to derive earlier terms.

Limit of a Sequence

The limit of a sequence is a concept used to determine what value a sequence approaches as the number of terms goes to infinity. In simpler words, it's about finding the value that the terms in the sequence get closer to as you continue down the sequence.
To find the limit of a sequence, we evaluate the behavior of its general term as \(n\) approaches infinity.
In our case, the general term is \(a_n = \frac{n}{2^{n+1}}\). As \(n\) increases, the numerator grows linearly while the denominator grows exponentially.
Exponential growth means that \(2^{n+1}\) grows much faster than \(n\). Thus, the fraction \(\frac{n}{2^{n+1}}\) becomes smaller and smaller, approaching zero.
Therefore, the limit of the sequence, \(\lim_{n \to \infty} a_n\), is 0.

Exponential Growth

Exponential growth is when a quantity increases rapidly by a consistent percentage over time. It is characterized by its multiplication rather than addition. This growth is contrasted with linear growth which increases by the same amount with each step.
In the context of sequences, examining the terms in the denominator \(2^{n+1}\) provides insight into how quickly a sequence grows or shrinks.
In our sequence, each increment in \(n\) results in the denominator multiplying by 2, thus growing exponentially. This characteristic of the denominator \(2^{n+1}\) is crucial for understanding the sequence's behavior as \(n\) increases.
Since exponential growth in the denominator substantially outweighs the linear growth in the numerator \(n\), the terms of the sequence become very small, showcasing the balancing act in computing sequence limits, resulting in convergence towards zero as observed.