Chapter 9: Problem 17
Write out the first five terms of the sequence, determine whether the sequence converges, and if so find its limit. $$ \left\\{\frac{(n+1)(n+2)}{2 n^{2}}\right\\}_{n=1}^{+\infty} $$
Short Answer
Expert verified
The sequence converges; the limit is \( \frac{1}{2} \).
Step by step solution
01
Write Out the First Five Terms
Given the sequence \( \left\{\frac{(n+1)(n+2)}{2n^2}\right\} \), we start by substituting the initial values of \( n \) from 1 to 5 to find the first five terms.- For \( n=1 \), the term is \( \frac{(1+1)(1+2)}{2 \cdot 1^2} = \frac{2 \times 3}{2} = 3 \).- For \( n=2 \), the term is \( \frac{(2+1)(2+2)}{2 \cdot 2^2} = \frac{3 \times 4}{8} = \frac{3}{2} \).- For \( n=3 \), the term is \( \frac{(3+1)(3+2)}{2 \cdot 3^2} = \frac{4 \times 5}{18} = \frac{10}{9} \).- For \( n=4 \), the term is \( \frac{(4+1)(4+2)}{2 \cdot 4^2} = \frac{5 \times 6}{32} = \frac{15}{16} \).- For \( n=5 \), the term is \( \frac{(5+1)(5+2)}{2 \cdot 5^2} = \frac{6 \times 7}{50} = \frac{21}{25} \).The first five terms are: 3, \( \frac{3}{2} \), \( \frac{10}{9} \), \( \frac{15}{16} \), \( \frac{21}{25} \).
02
Determine Convergence
To determine if the sequence converges, we calculate the limit as \( n \to \infty \) of the general term \( \frac{(n+1)(n+2)}{2n^2} \).First, expand the numerator: \[(n+1)(n+2) = n^2 + 3n + 2.\]Now we divide each term in the fraction by \( n^2 \):\[\frac{n^2 + 3n + 2}{2n^2} = \frac{n^2}{2n^2} + \frac{3n}{2n^2} + \frac{2}{2n^2} = \frac{1}{2} + \frac{3}{2n} + \frac{1}{n^2}.\]As \( n \to \infty \), the terms \( \frac{3}{2n} \) and \( \frac{1}{n^2} \) both approach 0. Therefore, the limit of the sequence is \( \frac{1}{2} \).
03
State the Conclusion
The sequence converges, and its limit is \( \frac{1}{2} \). We determined this by analyzing the limit of the formula for the sequence as \( n \) approaches infinity, finding that all but the constant term \( \frac{1}{2} \) vanish as \( n \) increases.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sequence Terms
A sequence is an ordered list of numbers, each called a term of the sequence. In mathematical notation, sequences are usually written with a general formula that describes the position of each term. Understanding the sequence terms is crucial when dealing with any mathematical sequence.
In our specific sequence, represented by \( \left\{ \frac{(n+1)(n+2)}{2n^2} \right\} \), each term is generated based on a value of \( n \).
To illustrate, let's calculate the first five terms:
In our specific sequence, represented by \( \left\{ \frac{(n+1)(n+2)}{2n^2} \right\} \), each term is generated based on a value of \( n \).
To illustrate, let's calculate the first five terms:
- For \( n = 1 \): The term is \( \frac{(1+1)(1+2)}{2 \cdot 1^2} = 3 \).
- For \( n = 2 \): The term is \( \frac{(2+1)(2+2)}{2 \cdot 2^2} = \frac{3}{2} \).
- For \( n = 3 \): The term is \( \frac{(3+1)(3+2)}{2 \cdot 3^2} = \frac{10}{9} \).
- For \( n = 4 \): The term is \( \frac{(4+1)(4+2)}{2 \cdot 4^2} = \frac{15}{16} \).
- For \( n = 5 \): The term is \( \frac{(5+1)(5+2)}{2 \cdot 5^2} = \frac{21}{25} \).
Limit of a Sequence
The concept of the limit of a sequence is pivotal in understanding whether a sequence converges—that is, whether it approaches a specific value as you examine more terms. The limit provides insight into the long-term behavior of a sequence.
In our exercise, we're interested in the behavior of the term \( \frac{(n+1)(n+2)}{2n^2} \) as \( n \to \infty \).
To find the limit, we expand the numerator, \( (n+1)(n+2) = n^2 + 3n + 2 \), and then simplify:
\[ \frac{n^2 + 3n + 2}{2n^2} = \frac{1}{2} + \frac{3}{2n} + \frac{1}{n^2}. \]
As \( n \to \infty \), the terms \( \frac{3}{2n} \) and \( \frac{1}{n^2} \) tend towards zero. What remains is the constant \( \frac{1}{2} \).
This analysis shows that no matter how large \( n \) gets, the sequence's terms get closer and closer to \( \frac{1}{2} \), thus the sequence converges to this limit.
In our exercise, we're interested in the behavior of the term \( \frac{(n+1)(n+2)}{2n^2} \) as \( n \to \infty \).
To find the limit, we expand the numerator, \( (n+1)(n+2) = n^2 + 3n + 2 \), and then simplify:
\[ \frac{n^2 + 3n + 2}{2n^2} = \frac{1}{2} + \frac{3}{2n} + \frac{1}{n^2}. \]
As \( n \to \infty \), the terms \( \frac{3}{2n} \) and \( \frac{1}{n^2} \) tend towards zero. What remains is the constant \( \frac{1}{2} \).
This analysis shows that no matter how large \( n \) gets, the sequence's terms get closer and closer to \( \frac{1}{2} \), thus the sequence converges to this limit.
Infinite Series
An infinite series involves summing the terms of a sequence without end. While closely related, it is a broader concept beyond our sequence's scope. The sequence in our example doesn't involve a summation from term to term but focuses on the limit of individual terms.
In cases where we would look at the infinite series related to a sequence, we'd be interested in whether the sum of its terms converges to a finite number.
Though not directly calculated in this exercise, considering the series formed by adding the terms of similar sequences over time helps in:
- Understanding convergence and divergence in another context.
- Applying these principles to complex problems in calculus and analysis.