Question:
Let {an}n≥1 be a sequence of real numbers such that a1+5m = 2, a2+5m = 3, a3+5m = 4, a4+5m = 5, a5+5m = 6, m = 0, 1, 2, … . Then limsupn→∞ an + liminfn→∞ an equals __________
Let {an}n≥1 be a sequence of real numbers such that a1+5m = 2, a2+5m = 3, a3+5m = 4, a4+5m = 5, a5+5m = 6, m = 0, 1, 2, … . Then limsupn→∞ an + liminfn→∞ an equals __________
Updated On: Nov 25, 2025
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Correct Answer: 8
Solution and Explanation
The given sequence {an} has a pattern based on a1+5m = 2, a2+5m = 3, a3+5m = 4, a4+5m = 5, a5+5m = 6 for any m ≥ 0. This indicates that the sequence is periodic with period 5. Therefore, the first 5 terms of each period are 2, 3, 4, 5, 6. To find lim sup and lim inf, consider the behavior of these periodic terms:
- limsupn→∞ an: This is the supremum of the limits of subsequences, equalling the maximum value in one period. Since the sequence is 2, 3, 4, 5, 6, maxi-mum is 6, so limsup is 6.
- liminfn→∞ an: This is the infimum of the limits of subsequences, equalling the minimum value in one period. Since the minimum is 2, liminf is 2.
Thus, limsupn→∞ an + liminfn→∞ an = 6 + 2 = 8.
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