Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{3 n+2}{n+1} $$

Step-by-Step Solution

Answer

First five terms: \(\frac{5}{2}, \frac{8}{3}, \frac{11}{4}, \frac{14}{5}, \frac{17}{6}\). The sequence converges to 3.

Step 1: Understand the Formula

We have the formula for the sequence: \(a_n = \frac{3n + 2}{n + 1}\). Our task is to find the first five terms and analyze the convergence of the sequence.

Step 2: Find the First Term

Substitute \(n = 1\) into the formula: \[a_1 = \frac{3(1) + 2}{1 + 1} = \frac{3 + 2}{2} = \frac{5}{2}\].

Step 3: Find the Second Term

Substitute \(n = 2\) into the formula: \[a_2 = \frac{3(2) + 2}{2 + 1} = \frac{6 + 2}{3} = \frac{8}{3}\].

Step 4: Find the Third Term

Substitute \(n = 3\) into the formula: \[a_3 = \frac{3(3) + 2}{3 + 1} = \frac{9 + 2}{4} = \frac{11}{4}\].

Step 5: Find the Fourth Term

Substitute \(n = 4\) into the formula: \[a_4 = \frac{3(4) + 2}{4 + 1} = \frac{12 + 2}{5} = \frac{14}{5}\].

Step 6: Find the Fifth Term

Substitute \(n = 5\) into the formula: \[a_5 = \frac{3(5) + 2}{5 + 1} = \frac{15 + 2}{6} = \frac{17}{6}\].

Step 7: Analyze the Sequence for Convergence

To determine if the sequence converges, analyze the limit \(\lim_{n \to \infty} \frac{3n + 2}{n + 1}\). This can be done by dividing the numerator and denominator by \(n\), resulting in \(\lim_{n \to \infty} \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}}\). As \(n\) approaches infinity, \(\frac{2}{n}\) and \(\frac{1}{n}\) approach 0.

Step 8: Calculate the Limit

The limit simplifies to \(\frac{3 + 0}{1 + 0} = 3\). Thus, the sequence converges to 3.

Key Concepts

Explicit Formula

An explicit formula allows us to calculate the terms of a sequence without needing to know the preceding term. In our case, the explicit formula given is \(a_n = \frac{3n + 2}{n + 1}\). This formula is a direct expression that helps calculate any term in the sequence instantly. Unlike a recursive formula, which requires previous term calculations, an explicit formula directly gives you the value of \(a_n\) once you substitute the value of \(n\). It is very convenient for deriving specific terms quickly. For instance, with this formula, finding the first few terms of the sequence is simply a matter of substituting the values of \(n\) like 1, 2, 3, etc.

Sequence Limit

The sequence limit is a crucial concept in understanding whether sequences converge or diverge. It refers to the value a sequence approaches as \(n\) becomes very large. To find the limit of the sequence given by \(a_n = \frac{3n + 2}{n + 1}\), we can simplify the expression by dividing the numerator and the denominator by \(n\). This gives us \(\lim_{n \to \infty} \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}}\). As \(n\) approaches infinity, the fractions \(\frac{2}{n}\) and \(\frac{1}{n}\) both go to zero, leaving the simplified expression \(\frac{3 + 0}{1 + 0} = 3\). Thus, the sequence converges to 3. Understanding sequence limits helps us predict long-term behavior and stability of sequences, which is essential in calculus and various applications.

First Terms of Sequence

Finding the first terms of a sequence gives us a clearer picture of how the sequence starts and behaves. With the formula \(a_n = \frac{3n + 2}{n + 1}\), we can calculate the first five terms:

• For \(n = 1\), \(a_1 = \frac{5}{2}\)
• For \(n = 2\), \(a_2 = \frac{8}{3}\)
• For \(n = 3\), \(a_3 = \frac{11}{4}\)
• For \(n = 4\), \(a_4 = \frac{14}{5}\)
• For \(n = 5\), \(a_5 = \frac{17}{6}\)

These terms help us to visualize the progression and trend of the sequence from the beginning. By identifying these early terms, you can get a sense if the sequence appears to be stabilizing towards a specific value, indicating convergence. First terms are also useful in checking calculations and understanding the function's behavior close to the origins of the sequence.