Question : Is the following true for any prime number $p\ge 5$ ?
If $\sum_{k=1}^{p-1}\frac{1}{k}=\frac ba$ where $\frac ba$ is an irreducible fraction, then $b$ can be divided by $p^2$.
Motivation : I've been interested in this summation. Then, I reached the above expectation. We can see this is true for smaller $p$.
$$\sum_{k=1}^{5-1}=\frac{5^2}{12},\sum_{k=1}^{7-1}=\frac{7^2}{20},\sum_{k=1}^{11-1}=\frac{{11}^2\cdot 61}{2520},\sum_{k=1}^{13-1}=\frac{{13}^2\cdot 509}{27720},\sum_{k=1}^{17-1}=\frac{{17}^2\cdot 8431}{720720}$$ So far I can neither find any counterexample even by using computer nor prove this is true for any prime number $p\ge 5$ ? Can anyone help?
This result is a corrollary of Wolstenholme's Theorem. The proof is well-documented.