Problem 44 - 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 find \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{(-1)^{n}}{n^{3}+3} $$

Step-by-Step Solution

Answer

The first five terms are \( \frac{1}{3}, \frac{-1}{4}, \frac{1}{11}, \frac{-1}{30}, \frac{1}{67} \). The limit is 0.

Step 1: Calculate the First Term

The first term of the sequence occurs when \( n = 0 \). Substitute \( n = 0 \) into \( a_n = \frac{(-1)^n}{n^3 + 3} \): \[ a_0 = \frac{(-1)^0}{0^3 + 3} = \frac{1}{3} \]

Step 2: Calculate the Second Term

The second term of the sequence occurs when \( n = 1 \). Substitute \( n = 1 \) into \( a_n = \frac{(-1)^n}{n^3 + 3} \): \[ a_1 = \frac{(-1)^1}{1^3 + 3} = \frac{-1}{4} \]

Step 3: Calculate the Third Term

The third term of the sequence occurs when \( n = 2 \). Substitute \( n = 2 \) into \( a_n = \frac{(-1)^n}{n^3 + 3} \): \[ a_2 = \frac{(-1)^2}{2^3 + 3} = \frac{1}{11} \]

Step 4: Calculate the Fourth Term

The fourth term of the sequence occurs when \( n = 3 \). Substitute \( n = 3 \) into \( a_n = \frac{(-1)^n}{n^3 + 3} \): \[ a_3 = \frac{(-1)^3}{3^3 + 3} = \frac{-1}{30} \]

Step 5: Calculate the Fifth Term

The fifth term of the sequence occurs when \( n = 4 \). Substitute \( n = 4 \) into \( a_n = \frac{(-1)^n}{n^3 + 3} \): \[ a_4 = \frac{(-1)^4}{4^3 + 3} = \frac{1}{67} \]

Step 6: List the First Five Terms

Now that we have computed each of the first five terms, the sequence is: \[ a_0 = \frac{1}{3}, \quad a_1 = \frac{-1}{4}, \quad a_2 = \frac{1}{11}, \quad a_3 = \frac{-1}{30}, \quad a_4 = \frac{1}{67} \]

Step 7: Determine the Limit as \(n\) Approaches Infinity

To find \( \lim_{n \to \infty} a_n \), consider the expression \( \frac{(-1)^n}{n^3 + 3} \). As \( n \) becomes very large, \( n^3 + 3 \) approaches infinity. Therefore, the expression approaches zero: \[ \lim_{n \to \infty} a_n = 0 \]

Key Concepts

Understanding the Limit of a Sequence

In mathematics, the limit of a sequence refers to the value that the terms of a sequence increasingly approach as the sequence progresses towards infinity. If a sequence \( \{a_n\} \) has a limit, it means that as we keep calculating more terms for higher values of \( n \), the terms come closer and closer to a specific number.
In this exercise, the given sequence is \( a_n = \frac{(-1)^n}{n^3 + 3} \). To find its limit, we observe what happens as \( n \) approaches infinity. The denominator \( n^3 + 3 \) becomes indefinitely large while the numerator oscillates between -1 and 1. Thus, the fraction's value trends towards zero.
This phenomenon can be explained simply: as the denominator grows significantly larger than the fixed values of the numerator, the overall fraction diminishes to zero. Hence, \( \lim_{n \to \infty} a_n = 0 \).

Getting the Hang of Term Calculation

Calculating terms of a sequence is basically plugging in different values of \( n \) to derive the terms. It's a straightforward process involving replacement and basic arithmetic calculations.
Let's use \( n = 0 \) as an initial example. By substituting \( n = 0 \) in \( a_n = \frac{(-1)^n}{n^3 + 3} \), we get \( a_0 = \frac{1}{3} \) because \( (-1)^0 = 1 \).
We perform similar substitutions for increasing values of \( n \):

• When \( n = 1 \, a_1 = \frac{-1}{4} \)
• When \( n = 2 \, a_2 = \frac{1}{11} \)
• When \( n = 3 \, a_3 = \frac{-1}{30} \)
• When \( n = 4 \, a_4 = \frac{1}{67} \)

By following this approach, each term of the sequence is revealed clearly, illustrating how each step embodies the progression of the sequence.

Exploring Alternating Sequences

An alternating sequence is one where the terms alternatively switch sign. This often occurs through the use of a factor like \( (-1)^n \). Here, each term in the sequence \( a_n = \frac{(-1)^n}{n^3 + 3} \) changes sign depending on the evenness or oddness of \( n \).
If \( n \) is even, \( (-1)^n \) equals 1, making the term positive. Conversely, if \( n \) is odd, \( (-1)^n \) equals -1, thus producing a negative term. This feature creates an interesting pattern in the behavior of sequences, affecting how we analyze them.
Such sequences can be found in various mathematical applications, providing valuable ways to model phenomena that exhibit alternating behaviors, such as vibrations, oscillations, or certain financial patterns.