Proving the sum of the harmonic series up to $p-1$ is divisible by $p$ Wolstenholmes theorem says that $1 + \frac{1}{2} + ... + \frac{1}{p-1} \equiv 0 \bmod p$. I dont quite get the proof, but I was wondering if the argument here is valid for the narrow case that it is divisible by p?
$$1 + \frac{1}{2} + \dots + \frac{1}{p-1} \equiv 1 + 2 + \dots + (p-1) \equiv \frac{p(p-1)}{2} \equiv 0 \bmod p$$
Regards (my first post on stackexchange :P)
 A: According to wikipedia the Wolstenholme theorem states that for $p>3$ the following two congruences hold:
$1+{1 \over 2}+{1 \over 3}+...+{1 \over p-1} \equiv 0 \pmod{p^2}$
$1+{1 \over 2^2}+{1 \over 3^2}+...+{1 \over (p-1)^2} \equiv 0 \pmod p.$
It goes on to explain:
"Congruences with fractions make sense, provided that the denominators are coprime to the modulus.
For example, with $p=7$, the first of these says that the numerator of $49/20$ is a multiple of $49$, while the second says the numerator of $5369/3600$ is a multiple of $7$."
You have a correct proof that for $p\geq 3$ the partial harmonic series viewed as integers mod $p$ is congruent to $0$ mod $p$.  As you-sir-33433 points out this actually is equivalent to saying that the partial harmonic series viewed as fractions has numerator congruent to $0$ mod $p$, by multiplying by common denominator $(p-1)!$  So you have correctly proven a statement which is weaker than the first congruency in Wolstenholme's theorem (but still relevant).  
A: For $n=1,2,...,(p-1)$ let $a_n = \frac{(p-1)!}{n}$.
Then making a common denominator:
$1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{p-1} = \frac{\sum_{n=1}^{(p-1)} a_n}{(p-1)!}.$
Now what you are trying to prove is that $p$ divides the numerator (since it definitely doesn't divide the denominator).
But this is easy since by Wilson's theorem $(p-1)! \equiv -1 \bmod p$, so $a_n \equiv -n^{-1} \bmod p$.
Thus $\sum_{n=1}^{(p-1)} a_n \equiv -\sum_{n=1}^{(p-1)} n^{-1} = -\sum_{m=1}^{(p-1)} m = \frac{p(p-1)}{2} \equiv 0 \bmod p$.
And so we are done...without having to consider the philosophy of "reducing fractions mod $p$".
