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MAT136H5 S 2025 - All Sections 5.6 preparation check
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Question: Is the series ∞ ∑ n=1 (n+2)! 1000n absolutely convergent, conditionally convergent, or divergent? A student submits the following solution: Line 1: In this example an= (n+2)! 1000n Line 2: To use the ratio test, we should evaluate the limit lim n→∞| an+1 an |, and then interpret the result. Line 3: lim n→∞| an+1 an |=lim n→∞| ( (n+3)! 1000n+1 ) ( (n+2)! 1000n ) | Line 4: =lim n→∞ (n+3)! 1000n+1 ⋅ 1000n (n+2)! Line 5: =lim n→∞ (n+2)!⋅(n+3)⋅1000n 1000n⋅1000⋅(n+2)! Line 6: =lim n→∞ n+3 1000 Line 7: =∞ Line 8: Because the limit is ∞, the ratio test tells us that the series ∞ ∑ n=1 (n+2)! 1000n is divergent. a) Is the above solution correct? [ Select ] No, it is not correct. Yes, it is correct. b) If it is incorrect, in which line is the first error? It is correct c) What is the correct final answer to the question? [ Select ] Conditionally convergent Absolutely convergent Divergent
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Step-by-Step Analysis
Carefully examine the problem statement and the proposed solution steps before judging each option.
Option a) 'Is the above solution correct?' with choices 'No, it is not correct.' and 'Yes, it is correct.'
- Consider the setup: the series is ∑ from n=1 to ∞ of (n+2)! / 1000^n. The student applies the ratio test by evaluating lim_{n→∞} |a_{n+1}/a_n| where a_n = (n+2)! / 1000^n. The ratio becomes
a_{n+1}/a_n = [(n+3)! / 1000^{n+1}] / [(n+2)! / 1000^n] = (n+3)/1000.
- As n grows, (n+3)/1000 grows without bound, so the limit is ∞, which is certainly greater than 1. By the ratio test, a limit greater than 1 implies the series diverges.
- The key check is whether this derivation is correct: the algebra leading to (n+3)/1000 is valid, and the conclusion that the limit is ∞ is correct.......Login to view full explanationLog in for full answers
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