调和级数的部分和不为整数

[randomatrices]

这个证明应该比下面那个连续合数定理要难点。

[TheMatrix]

这个证明应该比下面那个连续合数定理要难点。

部分和？连续的部分？

[randomatrices]

嗯， 从第一项开始的任意连续部分和。
我怀疑从任意项开始的任意连续部分和也不是整数，但我不知道怎么证。

部分和？连续的部分？

[san721]

嗯， 从第一项开始的任意连续部分和。
我怀疑从任意项开始的任意连续部分和也不是整数，但我不知道怎么证。

Suppose m>0. Assume the sum 1/(m+1)+1/(m+2)+...+1/(m+n) is an integer, then n>m. By Bertrand Postulate, there is a prime, say p, among these denominators. If 2p is not one of the denominators, then we stop. If 2p appears as a denominator, then by Bertrand postulate again, there is a prime number q>p appearing as a denominator... We end up with a prime number p\in (m, m+n] such that p is the only denominator that is divisible by p. Now, all terms 1/(m+k) are p-adic integers except for the term 1/p, this sum cannot be a rational integer.

[(ッ)]

假设1 + 1/2 + ... + 1/n = k

n! + n!/2 + ... + n!/n = n! * k

若n为质数，则
LHS = 1*2*...*(n-1) mod n，RHS = 0 mod n
若n不为质数, 记p为小于n的最大质数，根据博特兰-车比雪夫定理一定存在一个prime n/2 < p < n，这个p跟所有{2,3,..., n}互质
LHS = n! / p != 0 mod p, RHS = 0 mod p

[TheMatrix]

假设1 + 1/2 + ... + 1/n = k

n! + n!/2 + ... + n!/n = n! * k

若n为质数，则
LHS = 1*2*...*(n-1) mod n，RHS = 0 mod n
若n不为质数, 记p为小于n的最大质数，根据博特兰-车比雪夫定理一定存在一个prime n/2 < p < n，这个p跟所有{2,3,..., n}互质
LHS = n! / p != 0 mod p, RHS = 0 mod p

都用到了n<p<2n定理。不过你的是从1开始，san721是从任意数开始。

还可以再扩展一下原问题：调和级数中的任意有限子序列之和不为整数。这就难了吧？

[san721]

都用到了n<p<2n定理。不过你的是从1开始，san721是从任意数开始。

还可以再扩展一下原问题：调和级数中的任意有限子序列之和不为整数。这就难了吧？

This is not correct. For example, you can construct infinitely many subsequences each of which sums to 1.

1;
1/2+1/3+1/6;
1/2+1/3+1/7+1/42;
1/2+1/3+1/7+1/43+1/(42*43);
....

In fact, if you split the set of integers {n|n\geq 2} into two arbitrary subsets (more generally, any finitely many subsets), then it is proved that from one of these subsets there are infinitely many subsets of integers with the sums of the reciprocals are all integers. This was a question of Erdos (generalized from the so-call Egyptian Fraction question), solved by Ernie Croot in 1999.

[TheMatrix]

This is not correct. For example, you can construct infinitely many subsequences each of which sums to 1.

1;
1/2+1/3+1/6;
1/2+1/3+1/7+1/42;
1/2+1/3+1/7+1/43+1/(42*43);
....

In fact, if you split the set of integers {n|n\geq 2} into two arbitrary subsets (more generally, any finitely many subsets), then it is proved that from one of these subsets there are infinitely many subsets of integers with the sums of the reciprocals are all integers. This was a question of Erdos (generalized from the so-call Egyptian Fraction question), solved by Ernie Croot in 1999.

对啊！

这么一扩展，就有点知识体系的影子了。

[YWY]

假设1 + 1/2 + ... + 1/n = k

n! + n!/2 + ... + n!/n = n! * k

若n为质数，则
LHS = 1*2*...*(n-1) mod n，RHS = 0 mod n
若n不为质数, 记p为小于n的最大质数，根据博特兰-车比雪夫定理一定存在一个prime n/2 < p < n，这个p跟所有{2,3,..., n}互质
LHS = n! / p != 0 mod p, RHS = 0 mod p

都用到了n<p<2n定理。不过你的是从1开始，san721是从任意数开始。

还可以再扩展一下原问题：调和级数中的任意有限子序列之和不为整数。这就难了吧？

(ッ)的证明也适用于从m开始的情况，过程和方法和从1开始的情形完全一致。