Give an example of each or state that the request is impossible. For any that are impossible, give a compelling argument for why that is the case. a. A sequence with an infinite number of ones that does not converge to one. b. A sequence with an infinite number of ones that converges to a limit not equal to one. c. A divergent sequence such that for every n ∈ N it is possible to find n consecutive ones somewhere in the sequence.

See proofs below

Step-by-step explanation:

a) Consider the sequence {an}={(-1)^n}={-1,1,-1,1,...}

an has an infinite number of ones. Indeed, if n is even, (-1)^n=1, and there are infinite even numbers, hence infinite ones. However, {an} does not converge to one, because {an} is divergent (it oscillates between 1 and -1).

b) This is impossible. We will use the definition of limit to see why.

Let . Suppose that {an} has an infinite number of ones, and suppose that {an} converges to L. Then, there exists some M>0 such that if n≥M then . Now, there exist infinite natural numbers n≥M such that an=1 by hypotheses. Then

Hence L and 1 get arbitrarily close, therefore L=1.

c) It is possible to construct such a sequence:

{an}={1,0,1,1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,0,...}

even if there is not a simple explicit formula. The sequence is divergent because it has infinite zeroes and infinite ones (thus L should be either 0 or 1, but it can't be both).  

answer is:

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step-by-step explanation: