My coworker Paul Pedersen showed me another interesting observation/puzzle the other day:
I remember learning in high school that the harmonic series doesn’t converge:
1, 1+(1/2), 1+(1/2)+(1/3), 1+(1/2)+(1/3)+(1/4) …
There is a nice intuitive proof on Mathworld that dates back to the 14th century
So while 1 + (1/2) + (1/4) … never passes 2, the harmonic series reaches every number, but it grows very, very slowly, as you can see from this graph:

Paul had an interesting and surprising observation:
While the series passes an infinite number of integers at an increasingly slow rate, it never actually has an integer value. (Except for n=1)
For example it starts (3/2),(5/6),(13/12) just missing 2.
I’m still thinking about how to prove this fact. If you have a sum, (1/2) + (1/3) + … (1/n), you can reduce it to [(3*4*...*n)+(2*4*...*n)+...+(2*3*...*n-1)]/(2*3*…*n).
So in order for this to be an integer it has to be the case that for some integer N, [(3*4*...*n)+(2*4*...*n)+...+(2*3*...*n-1)] = (2*3*…*n)*N
We can rewrite this as [n*((n-1)*((n-2)*... (2+3) + 2*3) + 2*3*4) ... ) + 2*3*...*(n-1)] = (2*3*…*n)*N
We can call this n*A + 2*3*…*(n-1) = (2*3*…*n)*N.
If n is prime, we have the sum of a multiple of n and a non-multiple of n, and I think it’s clear that there is no N that will make this equation true.
In fact you have no chance of satisfying the equation until the 2n, since there will always be exactly one term missing the n factor and in that term no factor will be divisible by n.
Is it true that there will always be a prime between n and 2n? It seems like it should be true, since the density of prime numbers is on the order of (1/log n). I guess if someone has proved this then the proof is done, but based on the way Paul presented the problem to me, I think there might be a better way to show this…