The sequence defined by A is a sequence of real numbers, where each term is given by 32 - (-1)^n * 2n.

First, let's consider the two cases for n:

1. When n is even, (-1)^n = 1, so the nth term of the sequence is 32 - 2n.
2. When n is odd, (-1)^n = -1, so the nth term of the sequence is 32 + 2n.

Now, let's find the infimum (greatest lower bound) and supremum (least upper bound) of this sequence.

The infimum of A is the greatest number that is less than or equal to every number in the sequence. As n increases, the terms 32 - 2n (for even n) decrease without bound. Therefore, the infimum of A is negative infinity.

The supremum of A is the smallest number that is greater than or equal to every number in the sequence. As n increases, the terms 32 + 2n (for odd n) increase without bound. Therefore, the supremum of A is positive infinity.

So, inf A = -∞ and sup A = ∞.

Question

The sequence defined by A is a sequence of real numbers, where each term is given by 32 - (-1)^n * 2n.

First, let's consider the two cases for n:

1. When n is even, (-1)^n = 1, so the nth term of the sequence is 32 - 2n.
2. When n is odd, (-1)^n = -1, so the nth term of the sequence is 32 + 2n.

Now, let's find the infimum (greatest lower bound) and supremum (least upper bound) of this sequence.

The infimum of A is the greatest number that is less than or equal to every number in the sequence. As n increases, the terms 32 - 2n (for even n) decrease without bound. Therefore, the infimum of A is negative infinity.

The supremum of A is the smallest number that is greater than or equal to every number in the sequence. As n increases, the terms 32 + 2n (for odd n) increase without bound. Therefore, the supremum of A is positive infinity.

So, inf A = -∞ and sup A = ∞.

Knowee AI · Accepted Answer

The sequence defined by A is a sequence of real numbers, where each term is given by 32 - (-1)^n * 2n.

First, let's consider the two cases for n:

1. When n is even, (-1)^n = 1, so the nth term of the sequence is 32 - 2n.
2. When n is odd, (-1)^n = -1, so the nth term of the sequence is 32 + 2n.

Now, let's find the infimum (greatest lower bound) and supremum (least upper bound) of this sequence.

The infimum of A is the greatest number that is less than or equal to every number in the sequence. As n increases, the terms 32 - 2n (for even n) decrease without bound. Therefore, the infimum of A is negative infinity.

The supremum of A is the smallest number that is greater than or equal to every number in the sequence. As n increases, the terms 32 + 2n (for odd n) increase without bound. Therefore, the supremum of A is positive infinity.

So, inf A = -∞ and sup A = ∞.

LetA :=32 − (−1)n2n : n ∈ N.Find inf A and sup A. Justifications are required

Question

Let A := 32 − (−1)^n 2^n : n ∈ N. Find inf A and sup A. Justifications are required.

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LetA :=32 − (−1)n2n : n ∈ N.Find inf A and sup A. Justifications are required