$(p-1)^\text{th}$ Harmonic Number for Odd Prime $p$

As $p$ is an odd prime so there are $p-1$ i.e. even numbers of terms are there.Now we will start grouping as follows -

$\dfrac{1}{k} + \dfrac{1}{p-k}=\dfrac{p}{k(p-k)}$ where $0 \leq k \leq \dfrac{p-1}{2}$ .

The above sum can be rearranged as

$\dfrac{A_p}{B_p}=\displaystyle \sum_{k=0}^{\dfrac{p-1}{2}} \dfrac{p}{k(p-k)}$

The numerator hwill have a factor $p$ and it will not cancel out by any factor of denominator because there is no factor $p$ contains in it (as $p$ is a prime).

So, $A_p$ will be divisible by $p$ for every $p <50$ .