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Magazine Chapter 9: Problem 4
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 \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{3 n^{2}+2}{2 n-1} $$
Short Answer
Expert verified
The sequence diverges.
Step by step solution
01
Calculate the First Term
To find the first term of the sequence, substitute \(n=1\) into the explicit formula: \(a_1= \frac{3(1)^2 + 2}{2(1) - 1} = \frac{3\cdot1 + 2}{2 - 1} = \frac{5}{1} = 5\). So, the first term is 5.
02
Calculate the Second Term
For the second term, substitute \(n=2\) into the formula: \(a_2= \frac{3(2)^2 + 2}{2(2) - 1} = \frac{3\cdot4 + 2}{4 - 1} = \frac{14}{3}\). The second term is \(\frac{14}{3}\).
03
Calculate the Third Term
For the third term, substitute \(n=3\): \(a_3= \frac{3(3)^2 + 2}{2(3) - 1} = \frac{3\cdot9 + 2}{6 - 1} = \frac{29}{5}\). The third term is \(\frac{29}{5}\).
04
Calculate the Fourth Term
Substitute \(n=4\): \(a_4 = \frac{3(4)^2 + 2}{2(4) - 1} = \frac{3\cdot16 + 2}{8 - 1} = \frac{50}{7}\). The fourth term is \(\frac{50}{7}\).
05
Calculate the Fifth Term
Substitute \(n=5\): \(a_5 = \frac{3(5)^2 + 2}{2(5) - 1} = \frac{3\cdot25 + 2}{10 - 1} = \frac{77}{9}\). The fifth term is \(\frac{77}{9}\).
06
Determine Convergence or Divergence
To determine convergence, evaluate \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{3n^2 + 2}{2n - 1}\). Considering the highest powers of \(n\) in the numerator and denominator, the expression simplifies to \(\frac{3n^2}{2n} = \frac{3n}{2}\). Thus, as \(n\) approaches infinity, \(a_n\) approaches infinity. Therefore, the sequence diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Explicit Formula
An explicit formula is a mathematical expression that allows the direct computation of any term in a sequence. For our given sequence \(a_n = \frac{3n^2 + 2}{2n-1}\), the explicit formula makes it easy to calculate the value of any term \(a_n\) by simply substituting the desired term number \(n\) into the equation.
- To compute the first term \(a_1\) of this sequence, we substitute \(n = 1\) into the formula, which yields \(a_1 = 5\).
- For \(a_2\), substitute \(n = 2\) to get \(a_2 = \frac{14}{3}\).
- Continue by substituting now \(n = 3, 4,\) and \(5\) for subsequent terms.
- This method provides a straightforward approach to compute each of the first five terms specifically: 5, \(\frac{14}{3}\), \(\frac{29}{5}\), \(\frac{50}{7}\), and \(\frac{77}{9}\).
Limit of a Sequence
A fundamental aspect of sequences in mathematics is understanding their behavior as the term number \(n\) grows indefinitely. The limit of a sequence \(a_n\) is denoted as \(\lim_{n \to \infty} a_n\) and describes the value that the sequence approaches as \(n\) becomes infinitely large.
For the given sequence, we calculated:
For the given sequence, we calculated:
- \(\lim_{n \to \infty} \frac{3n^2 + 2}{2n - 1}\).
- To find this limit, look at the term with the highest degree of \(n\) in both the numerator and the denominator, which gives us \(\frac{3n^2}{2n}\).
- This simplifies to \(\frac{3n}{2}\).
- As \(n\) approaches infinity, \(\frac{3n}{2}\) also approaches infinity.
Sequence Convergence
In mathematical terms, a sequence is said to converge when its terms approach a specific value as \(n\) goes to infinity. If there's no such fixed value and the sequence keeps growing without bound, it diverges. To determine if a sequence converges, we primarily investigate its limit.
Consider the sequence \(a_n = \frac{3n^2 + 2}{2n - 1}\):
Consider the sequence \(a_n = \frac{3n^2 + 2}{2n - 1}\):
- The limit was found to be infinity, i.e., \(\lim_{n \to \infty} a_n = \infty\).
- Since the sequence continually increases without approaching any single finite value, this signifies divergence.
- A convergent sequence would have \(\lim_{n \to \infty} a_n = L\), where \(L\) would be a specific finite number.
Infinite Limit
An infinite limit occurs when the terms of a sequence grow beyond all bounds as \(n\) increases indefinitely. This means the sequence doesn't approach a finite value, leading to divergence instead of convergence.
For the sequence \(a_n = \frac{3n^2 + 2}{2n - 1}\), let's examine:
For the sequence \(a_n = \frac{3n^2 + 2}{2n - 1}\), let's examine:
- The calculation showed \(\lim_{n \to \infty} a_n\) is infinity.
- This result indicates that as \(n\) grows, \(a_n\) does not settle but keeps increasing.
- This type of sequence behavior is typical when the growth rate of the numerator is greater than that of the denominator in rational expressions.
