# Prove the sum of a harmonic series is never an integer

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• January 16th 2010, 05:06 PM
davismj
Prove the sum of a harmonic series is never an integer
http://i48.tinypic.com/mtt543.png

Not really sure how to do this. I thought about breaking it up into the numerator and denominator with a denominator of n!, but I don't know how to create a closed form expression of the numerator.

Either way, I don't think that's quite the right approach.
• January 16th 2010, 06:29 PM
tonio
Quote:

Originally Posted by davismj
http://i48.tinypic.com/mtt543.png

Not really sure how to do this. I thought about breaking it up into the numerator and denominator with a denominator of n!, but I don't know how to create a closed form expression of the numerator.

Either way, I don't think that's quite the right approach.

Let $k\in \mathbb{N}$ be s.t. $2^k\le n <2^{k+1}$ and let us define $L:=lcm\{2,3,...,n\}=2^k3^a5^b\cdot\ldots$ , so:

$\sum\limits_{i=2}^n\frac{1}{i} = \frac{(2^k3^a\cdot\ldots)\slash 2}{L}+\frac{(2^k3^a\cdot\ldots)\slash 3}{L}+\ldots +\frac{(2^k3^a\cdot\ldots)\slash n}{L}$

Now pay attention to the fact that all the numerators above are even except precisely one (why?), whereas the common denominator is even, and this mean this number can't be an integer...(Wink)

Tonio
• January 16th 2010, 06:46 PM
davismj
Quote:

Originally Posted by tonio
Let $k\in \mathbb{N}$ be s.t. $2^k\le n <2^{k+1}$ and let us define $L:=lcm\{2,3,...,n\}=2^k3^a5^b\cdot\ldots$ , so:

$\sum\limits_{i=2}^n\frac{1}{i} = \frac{(2^k3^a\cdot\ldots)\slash 2}{L}+\frac{(2^k3^a\cdot\ldots)\slash 3}{L}+\ldots +\frac{(2^k3^a\cdot\ldots)\slash n}{L}$

Now pay attention to the fact that all the numerators above are even except precisely one (why?)

Because when you get to the $(2^k-1)th$ term, $(2^k3^a\ldots)/2^k$ reduces to the product of the remaining primes, all odd.

Thank you muchly. I see it now.