Problem 57 - Calculus for Biology and Medicine

Question

Write the first five terms of the sequence \(\left\\{a_{n}\right\\}\) \(n=0,1,2,3, \ldots\), and determine whether \(\lim _{n \rightarrow \infty} a_{n}\) exists. If the limit exists, find it. $$ a_{n}=2^{n} $$

Step-by-Step Solution

Answer

The first five terms are 1, 2, 4, 8, 16. The limit does not exist as the sequence grows indefinitely.

Step 1: Identify the Sequence Formula

The sequence provided is defined by the formula \(a_n = 2^n\). Our task is to find the first five terms of this sequence by substituting the values of \(n\) from 0 to 4.

Step 2: Calculate the First Term

For \(n = 0\), substitute into the formula: \(a_0 = 2^0 = 1\). So, the first term \(a_0\) is 1.

Step 3: Calculate the Second Term

For \(n = 1\), substitute into the formula: \(a_1 = 2^1 = 2\). Thus, the second term \(a_1\) is 2.

Step 4: Calculate the Third Term

For \(n = 2\), substitute into the formula: \(a_2 = 2^2 = 4\). Therefore, the third term \(a_2\) is 4.

Step 5: Calculate the Fourth Term

For \(n = 3\), substitute into the formula: \(a_3 = 2^3 = 8\). This means the fourth term \(a_3\) is 8.

Step 6: Calculate the Fifth Term

For \(n = 4\), substitute into the formula: \(a_4 = 2^4 = 16\). Hence, the fifth term \(a_4\) is 16.

Step 7: Determine the Limit for Large \(n\)

To determine if the limit exists as \(n\) approaches infinity, analyze \(a_n = 2^n\). As \(n\) increases, \(2^n\) grows without bound. Thus, \(\lim_{n \to \infty} a_n\) does not exist because \(2^n\) goes to infinity.

Key Concepts

Exponential Sequence

An exponential sequence is a type of sequence where each term is a constant raised to the power of the term's position number. In mathematical terms, an exponential sequence can be represented as \(a_n = k^n\). Here, \(k\) is a positive constant and \(n\) is the position of the term, starting from 0. In the sequence provided in the exercise, \(k\) is 2. So, the sequence is defined as \(a_n = 2^n\).Exponential sequences have unique behaviors:

• The terms grow rapidly, especially if the base \(k\) is greater than 1.
• Starting from \(n = 0\), each successive term is obtained by multiplying the previous term by the base \(k\).

To illustrate, with \(a_n=2^n\), the first five terms are 1, 2, 4, 8, and 16. As you can see, each term doubles the previous one.

Limit of a Sequence

The limit of a sequence is an essential concept in mathematical analysis. It describes the behavior of the terms of the sequence as the position number \(n\) approaches infinity. If a sequence converges to a specific value, that value is called the limit of the sequence.For example, if a sequence \( \{b_n\} \) converges to \(L\), as \(n \to \infty\), then \( \lim_{n \to \infty} b_n = L \). When discussing limits, we are interested in whether the sequence approaches a finite value, remains constant, or diverges.For the exponential sequence \( a_n = 2^n \), as \( n \rightarrow \infty \), the terms grow exponentially without bound. Therefore, in this context, the sequence does not have a limit, emphasizing that not all sequences have finite limits.

Divergence of Sequence

A sequence diverges if it does not converge to a finite limit as \( n \to \infty \). Divergence means that the terms of the sequence continue to increase or decrease without approaching a specific value. In such cases, the sequence is said to diverge.The sequence \(a_n = 2^n\) is a classic example of divergence. As \(n\) increases, the term values become extraordinarily large, moving farther from any finite number. This behavior shows that the sequence continues to rise indefinitely.Characteristics of a divergent sequence include:

• The terms increase or decrease indefinitely without settling around a specific value.
• The sequence does not approach zero or any real number as \(n\) becomes very large.

Recognizing divergence is crucial, as it indicates that the sequence cannot be simplified to a numerical limit. Instead, it highlights the endless growth or decay of its terms.