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 enoughthat:
∑k
1 1 1 1 = + + . . . > Mn 1 2 3
n =1
∑
∞
Note that this statement means that the infinite harmonic series, , grows without 1
n =1
bound, or diverges. For the specific example M = 5, find a value of k , not necessarily thensmallest , such that the inequality holds. Justify your answer.
英文原题
Prove that the product of two juicy numbers (not necessarily distinct) is always a juicy
number. Hint: if j 1 and j 2 are the two numbers, how can you change the decompositions of
1 ending in 1
j 1 or 1
j 2 to make them end in 1
j 1 j 2 ?
2
解析
英文解析
- 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 enoughthat:
∑k
1 1 1 1 = + + . . . > Mn 1 2 3
n =1
∑
∞
Note that this statement means that the infinite harmonic series, , grows without 1
n =1
bound, or diverges. For the specific example M = 5, find a value of k , not necessarily thensmallest , such that the inequality holds. Justify your answer.
1 1 1 1 1 1
Solution: Note that + + . . . + > + . . . + = . Therefore, if we apply thisn +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 + + + + + + + . . . + + . . . + > + . . . + = 4
2 3 4 5 6 7 8 129 256 2 2
so, adding in , we get 1
2561
∑
51
n =1 nso k = 256 will suffice.