HMMT 十一月 2008 · 团队赛 · 第 5 题
HMMT November 2008 — Team Round — Problem 5
题目详情
- An aside: the sum of all the unit fractions It is possible to show that, given any real M, there exists a positive integer k large enough that: k ∑ 1 1 1 1 = + + . . . > M n 1 2 3 n =1 ∑ ∞ 1 Note that this statement means that the infinite harmonic series, , grows without n =1 n bound, or diverges. For the specific example M = 5, find a value of k , not necessarily the smallest , such that the inequality holds. Justify your answer.
解析
- An aside: the sum of all the unit fractions It is possible to show that, given any real M, there exists a positive integer k large enough that: k ∑ 1 1 1 1 = + + . . . > M n 1 2 3 n =1 ∑ ∞ 1 Note that this statement means that the infinite harmonic series, , grows without n =1 n bound, or diverges. For the specific example M = 5, find a value of k , not necessarily the smallest , such that the inequality holds. Justify your answer. 1 1 1 1 1 1 Solution: Note that + + . . . + > + . . . + = . Therefore, if we apply this n +1 n +2 2 n 2 n 2 n 2 to n = 1 , 2 , 4 , 8 , 16 , 32 , 64 , 128, we get ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 1 1
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- . . . + + . . . + > + . . . + = 4 2 3 4 5 6 7 8 129 256 2 2 1 so, adding in , we get 1 256 ∑ 1
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5 n n =1 so k = 256 will suffice.