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Chapter 3: Problem 64

For the sequence b defined by \(b_{n}=n(-1)^{n}, n \geq 1\). Is \(b\) nondecreasing?

Short Answer

Expert verified
The sequence \(b_{n}=n(-1)^{n}\) is not nondecreasing.

Step by step solution

01

Analyze the sequence for even values of \(n\)

For even values of \(n\), the equation becomes \(b_{n}=n(-1)^{n}=n(-1)^{2k} = n*(1) = n\). Incrementing \(n\) by 1 gives \(b_{n+1}=(n+1)(-1)^{n+1} = (n+1)(-1) = -(n+1)\). We can see that \(b_{n+1} < b_{n}\), because \(-(n+1) < n\) for all \(n > 0\). Therefore, the sequence \(b_{n}\) is decreasing when \(n\) is even.
02

Analyze the sequence for odd values of \(n\)

This time we consider when \(n\) is odd. The equation becomes \(b_{n}=n(-1)^{n}= n(-1)^{2k+1} = n*(-1) = -n\). Incrementing \(n\) by 1 gives \(b_{n+1}=(n+1)(-1)^{n+1} = (n+1)(1) = n+1\). Here, \(b_{n+1} > b_{n}\), because \(n+1 > -n\) for all \(n > 0\). Hence, the sequence \(b_{n}\) is increasing when \(n\) is odd.
03

Conclusion

Because the sequence \(b_{n}\) is decreasing when \(n\) is even and increasing when \(n\) is odd, one can conclude that the sequence is not nondecreasing over all \(n\). Nondecreasing requires that the sequence be either constant or increasing for all \(n\), and in this case, that condition is not satisfied.

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Key Concepts

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

Nondecreasing Sequence
A sequence is called nondecreasing if each term is at least as large as the previous one. This means for a sequence \(a_n\), it satisfies \(a_{n+1} \geq a_n\) for all \(n\). Nondecreasing sequences can also remain constant or increase at different intervals. However, they must never decrease.
Consider the sequence \(b_n = n(-1)^n\), which has alternating signs due to the \((-1)^n\) term. For even \(n\), \(b_n\) takes positive values; for odd \(n\), \(b_n\) takes negative values.
  • When \(n\) is even, \(b_n = n\) and the next term \(b_{n+1} = -(n+1)\) causes a drop in the sequence.
  • When \(n\) is odd, \(b_n = -n\), and the next term \(b_{n+1} = n+1\) results in an increase.
Hence, \(b_n\) cannot be nondecreasing overall since it fails to maintain a consistent direction without dropping.
Mathematical Proof
Mathematical proof is a logical argument that verifies the truth of a statement. Proofs can take various forms, such as direct, indirect, or by contradiction. In verifying sequences, proving properties like increasing, decreasing, or constant behavior is essential.
For the given sequence \(b_n = n(-1)^n\), we implemented a direct analysis by checking the behavior separately for even and odd \(n\):
  • For even \(n\), we derived \(b_{n+1} < b_n\), proving that the sequence decreases.
  • For odd \(n\), we established \(b_{n+1} > b_n\), confirming an increase.
Combining these observations, we arrived at our conclusion that \(b_n\) is not nondecreasing, as was required by the proof.
Odd and Even Functions
The concept of odd and even functions intertwines with sequences since they heavily affect patterns. Functions are named odd or even based on their symmetry properties.
  • An even function \(f(x)\) satisfies \(f(x) = f(-x)\). It reflects symmetry around the y-axis.
  • An odd function \(f(x)\) meets \(f(-x) = -f(x)\). It shows symmetry around the origin.
In our sequence, the term \((-1)^n\) plays a key role in alternating each term's sign, producing behavior akin to these function properties. For even \(n\), the term equates to one, aligning with its nature. For odd \(n\), it equals negative one, matching odd functions' transformative properties. Understanding this integration helps dissect sequences better.

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Most popular questions from this chapter

List all strings over \(X=\\{0,1\\}\) of length 2 .

Prove that for all real numbers \(x\) and integers \(n,\lceil x\rceil=\) \(n\) if and only if there exists \(\varepsilon, 0 \leq \varepsilon<1\), such that \(x+\varepsilon=n\)

If \(X\) and \(Y\) are sets, we define \(X\) to be equivalent to \(Y\) if there is a one-to-one, onto function from \(X\) to \(Y .\) Show that for any set \(X, X\) is not equivalent to \(\mathcal{P}(X),\) the power \(\operatorname{set}\) of \(X\).

Let \(L\) be the set of all strings, including the null string, that can be constructed by repeated application of the following rules: If \(\alpha \in L,\) then \(a \alpha b \in L\) and \(b \alpha a \in L .\) If \(\alpha \in L\) and \(\beta \in L,\) then \(\alpha \beta \in L\) For example, \(a b\) is in \(L\), for if we take \(\alpha=\lambda,\) then \(\alpha \in L\) and the first rule states that \(a b=a \alpha b \in L\). Similarly, \(b a \in L\). As another example, aabb is in \(L\), for if we take \(\alpha=a b\), then \(\alpha \in L\); by the first rule, \(a a b b=a \alpha b \in L .\) As a final example, aabbba is in \(L\), for if we take \(\alpha=a a b b\) and \(\beta=b a,\) then \(\alpha \in L\) and \(\beta \in L ;\) by the second rule, \(a a b b b a=\alpha \beta \in L\). Show that \(a a b\) is not in \(L\).

Find \(b_{n}, n=1, \ldots, 6,\) where $$ b_{n}=n+(n-1)(n-2)(n-3)(n-4)(n-5) $$
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