Problem 4
Question
Determine whether the sequence \(\left\\{a_{n}\right\\}\) converges. If it does, state the limit. $$ a_{n}=3-(-1)^{n} $$
Step-by-Step Solution
Verified Answer
The sequence does not converge.
1Step 1: Identify the Pattern in the Sequence
Examine the sequence formula given by \( a_n = 3 - (-1)^n \). The term \((-1)^n\) indicates an alternating sequence because it takes the value of \(-1\) when \(n\) is odd and \(1\) when \(n\) is even.
2Step 2: Analyze Terms for Even and Odd n
For even \(n\), \((-1)^n = 1\), so the sequence term becomes \(a_n = 3 - 1 = 2\). For odd \(n\), \((-1)^n = -1\), so the sequence term becomes \(a_n = 3 - (-1) = 4\).
3Step 3: Determine the Terms of the Sequence
We have determined that the terms alternate between \(2\) and \(4\). Thus, the sequence \(a_n\) is: \(2, 4, 2, 4, 2, 4, \ldots\).
4Step 4: Evaluate Convergence
A sequence converges if the terms approach a single value as \(n\) goes to infinity. Here, the terms do not approach a single value; instead, they alternate between \(2\) and \(4\).
5Step 5: Conclusion
Since the sequence does not settle to a single limit value, the sequence \(\{a_n\}\) does not converge.
Key Concepts
Alternating SequenceSequence LimitsConvergence Evaluation
Alternating Sequence
Sequences are fascinating and alternating sequences bring an extra layer of intrigue. In mathematics, an alternating sequence is a sequence whose terms switch signs as the sequence progresses. A common indicator is found within the sequence formula — using an expression like \((-1)^n\), where the sign changes based on whether n is odd or even.
For instance, in the sequence formula given in our example,\(a_n = 3 - (-1)^n\),we observe that \((-1)^n\)induces this alternating behavior:
For instance, in the sequence formula given in our example,\(a_n = 3 - (-1)^n\),we observe that \((-1)^n\)induces this alternating behavior:
- When n is even, \((-1)^n = 1\),and the sequence value becomes 2.
- When n is odd, \((-1)^n = -1\),and the sequence value becomes 4.
Sequence Limits
The concept of a sequence limit is pivotal in understanding sequence convergence. A sequence limit is a value that terms of the sequence "approach" as the index n becomes very large — moving toward infinity. When a sequence has a limit, all its terms get arbitrarily close to this value.
However, in the case of our sequence, \(a_n = 3 - (-1)^n\),the terms are clearly alternating between two specific numbers: 2 and 4. This indicates that:
However, in the case of our sequence, \(a_n = 3 - (-1)^n\),the terms are clearly alternating between two specific numbers: 2 and 4. This indicates that:
- There is no single fixed point or number that these terms converge to, as they do not hone in on one particular value.
- The definition of a sequence limit requires that as n approaches infinity, the sequence terms must approach one specific fixed value.
Convergence Evaluation
Evaluating the convergence of a sequence requires assessing whether the terms collectively approach a particular value as the sequence progresses to infinity. This is known as convergence.
For the sequence described by \(a_n = 3 - (-1)^n\),convergence evaluation involves examining the terms as given:
For the sequence described by \(a_n = 3 - (-1)^n\),convergence evaluation involves examining the terms as given:
- Even indexed terms are consistently 2
- Odd indexed terms are consistently 4
- There is no indication of terms approaching a singular "limit" value.
- The lack of a single point toward which the sequence tends means that the sequence does not exhibit convergence.
Other exercises in this chapter
Problem 4
In Exercises \(1-8\), evaluate the given limit. $$ \lim _{x \rightarrow 12}(3-x / 4) $$
View solution Problem 4
Simplify the given expression. $$ \left(8^{\sqrt{3}} \cdot 4^{\sqrt{7}}\right) / 2^{\pi} $$
View solution Problem 5
Decide whether the indicated limit exists. If the limit does exist, compute it. $$ \lim _{h \rightarrow 1} \frac{h-3}{h+1} $$
View solution Problem 5
Determine the values at which the given function \(f\) is continuous. Remember that if \(c\) is not in the domain of \(f,\) then \(f\) cannot be continuous at \
View solution