Question

Claim: If (s_{n}) is any sequence of real numbers with
?_{?+1} = ?_{?}^{2} + 3?_{?} for
all n in N, then ?_{?} ≥ 0 for all n in N.

Proof: Suppose (s_{n}) is any sequence of real numbers
with ?_{?+1} = ?_{?}^{2} + 3?_{?}
for all n in N. Let P(n) be the inequality statements ?_{?}
≥ 0.

Let k be in N and suppose P(k) is true: Suppose ?_{?} ≥
0.

Note that ?_{?+1} = ?_{?}^{2} +
3?_{?} = ?_{?} (?_{?} + 3). Adding 3 to
both sides of the inequality ?_{?} ≥ 0 yields
(?_{?} + 3) ≥ 0 + 3 ≥ 0.

Multiplying both sides of the inequality (?_{?} + 3) ≥ 0 by
?_{?} yields ?_{?} (?_{?} + 3) ≥
?_{?} ∙ 0 with the ≥ direction preserved since
?_{?} ≥ 0.

?_{?+1} ≥ 0 since ?_{?+1}= ?_{?}
(?_{?} + 3). So P(k + 1) is true.

We conclude that P(n) is true for all n in N.

(a) Is the Claim true or false? Explain carefully if the Claim is false. (no additional explanation if true.)

(b) Critique the proof. Is it complete? Does it prove the claim? What are the flaws, if any? Be specific, identifying all flaws.

Answer #1

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ASSUMPTIONS
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positive integer N such that for all n ≥
N, |an – A| < ε .
Let P be 6. and Let Q be 24.
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sequence.
ASSUMPTIONS
Definition: A sequence {an} for n = 1 to ∞ converges
to a real number A if and only if for each ε > 0 there is a
positive integer N such that for all n ≥
N, |an – A| < ε .
Let P be 6. and Let Q be 24.
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