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标题: 把自然数表示为自然数倒数的和，行不？ [打印本页]

作者: 新用户 时间: 2006-11-23 01:01
标题: 把自然数表示为自然数倒数的和，行不？

来源： pistons

问题: 能否把一个自然数n写成一些(有限个)互不相等的自然数的倒数的和?
举例:
n=1: 1=1/1;
n=2: 2=1/1+1/2+1/3+1/6

那么n=3行不行? n=4行不行?具体怎么表示?
n是一般的自然数呢?

作者: 未经注册笔名 时间: 2006-11-23 05:53
标题: yes!

1 = 1/2 + 1/3 + 1/6, based on this:

for any k, we can write 1 = 1/K_1 + ... + 1/K_r, for some K_1, ..., K_r, such that k < K_1 < K_2 < K_3 < ... < K_r.

作者: 未经注册笔名 时间: 2006-11-23 06:31
标题: Explanation above may mislead some one.

Let me give more explanation.

We know that 1/1 + 1/2 + ... + 1/n + ... diverges to +OO.

Thus for any positive integer k, we have

1/(k+1) + 1/(k+2) + ... + 1/n + ... diverges to +OO.

So there exists an integer r, such that

1/(k+1) + ... + 1/(k+r) < 1 <= 1/(k+1) + ... + 1/(k+r) + 1/(k+r+1).

If the equation holds, we are done.

Otherwise,

Let a/b = 1 - 1/(k+1) - 1/(k+2) - ... - 1/(k+r), a and b are co-prime.

We know that 0 < a/b < 1/(k+r+1).

If a = 1, we are done since obviously b > k + r + 1
Otherwise, b = a*q + r where 0 < r < a.

Then 1/(q+1) < a/b. So q+1 > k+r+1.

a/b - 1/(q+1) = (aq + a - b)/(qb + b).

0 < aq + a - b = a - r < a.

If we simpilfy a/b - 1/(q+1), we get a rational number with a numerator smaller than a, and a denominator bigger than b.

If we repeat the steps above until the numerator becomes 1.

Since the numerator is a strict decreasing sequence, the repition will stop after a finite number of steps.

For example, let k = 1.

we have 1/2 + 1/3 < 1 < 1/2 + 1/3 + 1/4.

1 - 1/2 - 1/3 = 1/6

So 1 = 1/2 + 1/3 + 1/6

作者: 未经注册笔名 时间: 2006-11-23 06:39
标题: Another one:

1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/230 + 1/57960 = 1.

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