Problem 2
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 \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{3 n+2}{n+1}\)
Step-by-Step Solution
Verified Answer
The sequence converges to 3.
1Step 1: Identify the First Five Terms
We start by finding the first five terms of the sequence using the given explicit formula \( a_n = \frac{3n + 2}{n + 1} \). Substitute \( n = 1, 2, 3, 4, 5 \) into the formula. - For \( n = 1 \): \( a_1 = \frac{3 imes 1 + 2}{1 + 1} = \frac{5}{2} = 2.5 \)- For \( n = 2 \): \( a_2 = \frac{3 imes 2 + 2}{2 + 1} = \frac{8}{3} \approx 2.67 \)- For \( n = 3 \): \( a_3 = \frac{3 imes 3 + 2}{3 + 1} = \frac{11}{4} = 2.75 \)- For \( n = 4 \): \( a_4 = \frac{3 imes 4 + 2}{4 + 1} = \frac{14}{5} = 2.8 \)- For \( n = 5 \): \( a_5 = \frac{3 imes 5 + 2}{5 + 1} = \frac{17}{6} \approx 2.83 \) Thus, the first five terms are approximately 2.5, 2.67, 2.75, 2.8, and 2.83.
2Step 2: Analyze Convergence or Divergence
To determine if the sequence converges, consider the behavior of \( a_n \) as \( n \) approaches infinity. Factor \( n \) out of the numerator and denominator: \[ a_n = \frac{3n + 2}{n + 1} = \frac{n(3 + \frac{2}{n})}{n(1 + \frac{1}{n})} = \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}} \]As \( n \to \infty \), \( \frac{2}{n} \to 0 \) and \( \frac{1}{n} \to 0 \), so \[ \lim_{n \to \infty} \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}} = \frac{3 + 0}{1 + 0} = 3 \]Therefore, the sequence converges to 3.
3Step 3: Conclusion
Since the limit of \( a_n \) as \( n \) approaches infinity is 3, the sequence \( \{a_n\} \) converges to 3. Therefore, the solution to the problem is that the sequence converges, and \( \lim_{n \to \infty} a_n = 3 \).
Key Concepts
Convergence of SequencesLimit of a SequenceExplicit Formula for a Sequence
Convergence of Sequences
In mathematics, understanding whether a sequence converges or diverges is crucial. A sequence converges when its terms get closer and closer to a specific value as the sequence progresses towards infinity. Conversely, a sequence diverges if it does not approach any particular value.
For example, looking at the sequence given by the explicit formula \( a_n = \frac{3n + 2}{n + 1} \), we want to explore its behavior as \( n \) increases. When examining the first few terms—2.5, 2.67, 2.75, 2.8, and 2.83—we see that they appear to be approaching a specific value. This hints at convergence, but we must prove it analytically.
To do this, algebraic manipulation of the formula can help determine if the sequence converges. Simplifying to \( \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}} \) allows us to see how it behaves as \( n \to \infty \). As small fractions like \( \frac{2}{n} \) and \( \frac{1}{n} \) tend to zero, we confirm the terms are approaching the limit, suggesting convergence to 3.
For example, looking at the sequence given by the explicit formula \( a_n = \frac{3n + 2}{n + 1} \), we want to explore its behavior as \( n \) increases. When examining the first few terms—2.5, 2.67, 2.75, 2.8, and 2.83—we see that they appear to be approaching a specific value. This hints at convergence, but we must prove it analytically.
To do this, algebraic manipulation of the formula can help determine if the sequence converges. Simplifying to \( \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}} \) allows us to see how it behaves as \( n \to \infty \). As small fractions like \( \frac{2}{n} \) and \( \frac{1}{n} \) tend to zero, we confirm the terms are approaching the limit, suggesting convergence to 3.
Limit of a Sequence
The concept of a limit in sequences is essential for determining the behavior of terms as they progress towards infinity. The limit of a sequence is the value that the terms approach as the index, in our case \( n \), goes to infinity.
In our example, the sequence defined as \( a_n = \frac{3n + 2}{n + 1} \) reaches its limit evaluated as follows: factor \( n \) from the numerator and denominator: \( \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}} \). When \( n \) becomes infinitely large, the tiny contributions \( \frac{2}{n} \) and \( \frac{1}{n} \) diminish to zero. This simplifies our expression to \( \frac{3}{1} = 3 \).
In our example, the sequence defined as \( a_n = \frac{3n + 2}{n + 1} \) reaches its limit evaluated as follows: factor \( n \) from the numerator and denominator: \( \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}} \). When \( n \) becomes infinitely large, the tiny contributions \( \frac{2}{n} \) and \( \frac{1}{n} \) diminish to zero. This simplifies our expression to \( \frac{3}{1} = 3 \).
- This computation demonstrates that the upper-bound value the sequence terms approach is 3.
- Knowing the limit helps propose conclusions about the nature of the sequence, offering critical insights into its behavior and long-term trends.
Explicit Formula for a Sequence
An explicit formula allows us to directly compute the terms of a sequence without relying on previous terms. This formula gives a clear rule for finding each term, typically involving the term's index, \( n \).
For example, the sequence defined by the explicit formula \( a_n = \frac{3n + 2}{n + 1} \) makes it easy to generate terms by substituting different values of \( n \).
For example, the sequence defined by the explicit formula \( a_n = \frac{3n + 2}{n + 1} \) makes it easy to generate terms by substituting different values of \( n \).
- Using \( n = 1, 2, 3, 4, 5 \) produces the terms 2.5, 2.67, 2.75, 2.8, and 2.83, respectively.
- Having an explicit formula benefits analysis, as it can readily demonstrate the behavior and limit of the sequence.
Other exercises in this chapter
Problem 1
In Problems 1-14, indicate whether the given series converges or diverges. If it converges, find its sum. Hint: It may help you to write out the first few terms
View solution Problem 2
In Problems \(1-8\), find the convergence set for the given power series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{3^{n}} $$
View solution Problem 2
In Problems 1-18, find the terms through \(x^{5}\) in the Maclaurin series for \(f(x)\). Hint: It may be easiest to use known Maclaurin series and then perform
View solution Problem 2
\(\sum_{k=1}^{\infty} \frac{3}{2 k-3}\)
View solution