Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{n^{3}+3 n^{2}+3 n}{(n+1)^{3}}\)

Step-by-Step Solution

Answer

The sequence converges and \( \lim_{n \to \infty} a_n = 1 \).

Step 1: Understand the Explicit Formula

The explicit formula given is \( a_{n} = \frac{n^3 + 3n^2 + 3n}{(n+1)^3} \). This formula represents a sequence where each term is defined in terms of \( n \). We are tasked to find the first five terms and analyze convergence.

Step 2: Calculate the First Five Terms

We need to substitute \( n = 1, 2, 3, 4, 5 \) into the expression to find \( a_1, a_2, a_3, a_4, \) and \( a_5 \).- \( a_1 = \frac{1^3 + 3 \times 1^2 + 3 \times 1}{(1+1)^3} = \frac{1 + 3 + 3}{8} = \frac{7}{8} \)- \( a_2 = \frac{2^3 + 3 \times 2^2 + 3 \times 2}{(2+1)^3} = \frac{8 + 12 + 6}{27} = \frac{26}{27} \)- \( a_3 = \frac{3^3 + 3 \times 3^2 + 3 \times 3}{(3+1)^3} = \frac{27 + 27 + 9}{64} = \frac{63}{64} \)- \( a_4 = \frac{4^3 + 3 \times 4^2 + 3 \times 4}{(4+1)^3} = \frac{64 + 48 + 12}{125} = \frac{124}{125} \)- \( a_5 = \frac{5^3 + 3 \times 5^2 + 3 \times 5}{(5+1)^3} = \frac{125 + 75 + 15}{216} = \frac{215}{216} \)

Step 3: Determine Convergence or Divergence

To determine if the sequence converges or diverges, we must examine the behavior of \( a_n \) as \( n \to \infty \). We observe that the expression \( \frac{n^3 + 3n^2 + 3n}{(n+1)^3} \) is dominated by \( n^3 \) in both the numerator and denominator as \( n \) becomes very large. Simplifying gives:\[ a_n = \frac{n^3 + 3n^2 + 3n}{(n+1)^3} \approx \frac{n^3}{n^3} = 1 \]Thus, the sequence converges to 1.

Step 4: Find the Limit as \(n \to \infty\)

Since the dominant term \( n^3 \) in the numerator and denominator cancels out as \( n \) increases, the limit of \( a_n \) as \( n \to \infty \) is:\[ \lim_{n \to \infty} a_n = 1 \]

Key Concepts

Explicit Formula

An explicit formula in sequences is a way to define a sequence where the terms are given directly in terms of the position number, like \( n \). In the formula \( a_n = \frac{n^3 + 3n^2 + 3n}{(n+1)^3} \), each term is calculated based on the number \( n \). By substituting different values of \( n \), such as 1, 2, 3, and so on, we can easily calculate the terms of the sequence.

• For \( n = 1 \), the term is \( \frac{7}{8} \).
• For \( n = 2 \), it is \( \frac{26}{27} \).
• For \( n = 3 \), we get \( \frac{63}{64} \).
• For \( n = 4 \), it becomes \( \frac{124}{125} \).
• For \( n = 5 \), the term is \( \frac{215}{216} \).

Using an explicit formula helps us quickly determine the values of a sequence without needing the previous terms. This makes it a handy tool in mathematics for efficiently analyzing sequences and spotting patterns.

Limit of a Sequence

The limit of a sequence is what the terms of the sequence approach as the position number \( n \) becomes very large. It helps to determine the behavior of a sequence over the long term. In mathematical terms, it's written as \( \lim_{n \to \infty} a_n \). When calculating this, we are interested in what value \( a_n \) approaches as \( n \) goes to infinity.For the formula \( a_n = \frac{n^3 + 3n^2 + 3n}{(n+1)^3} \), we consider the dominant term for large \( n \). Both the numerator and the denominator mainly consist of the \( n^3 \) term. This simplifies to:\[\lim_{n \to \infty} a_n = \frac{n^3}{n^3} = 1\]This shows that as \( n \) increases, the terms get closer and closer to 1. Finding the limit gives us useful information about the expected value when the sequence continues indefinitely.

Convergence and Divergence of Sequences

Sequences can either converge or diverge as \( n \) increases.

• A sequence converges if it settles into a specific value as \( n \) approaches infinity.
• Conversely, it diverges if it does not settle to any particular value or grows without bound.

To see if our sequence converges, look at \( a_n = \frac{n^3 + 3n^2 + 3n}{(n+1)^3} \). As we observed, simplifying for large \( n \) shows that the expression approximates 1:\[\frac{n^3 + 3n^2 + 3n}{(n+1)^3} \approx 1\]Therefore, this sequence converges, and the limit is 1. Identifying whether a sequence converges or diverges is crucial for understanding its long-term behavior. Convergence implies stability, while divergence points to unbounded growth or erratic behavior.