/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Determine whether the sequence \... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the sequence \(\left\\{a_{n}\right\\}\) converges. If it does, state the limit. $$ a_{n}=3-(-1)^{n} $$

Short Answer

Expert verified
The sequence does not converge.

Step by step solution

01

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.
02

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 \(a_n = 3 - (-1) = 4\).
03

Determine the Terms of the Sequence

We have determined that the terms alternate between \(2\) and \(4\). Thus, the sequence \(a_n\) is: \(2, 4, 2, 4, 2, 4, \ldots\).
04

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 \(4\).
05

Conclusion

Since the sequence does not settle to a single limit value, the sequence \(\{a_n\}\) does not converge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • 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.
Thus, the sequence consistentizes as: 2, 4, 2, 4, and so on. Alternating sequences have terms that do not just vary in numerical value but switch signs or values back and forth, making them non-monotonic by nature.
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:
  • 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.
In our example, because the values just switch back and forth between two numbers restrictively, it fails to meet the definition of sequence limits and therefore, does not converge.
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:
  • Even indexed terms are consistently 2
  • Odd indexed terms are consistently 4
Because the terms oscillate, we see:
  • 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.
In simpler terms, convergence implies security at one value as n goes infinite, which does not happen in an alternating pattern like this. Thus, our sequence lacks convergence due to its alternating nature, remaining a jumping act between two distinct values.

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