The harmonic series of prime numbers can be written as $\sum_{i=1}^{n}\frac{1}{p_{i}} = \frac{p_{2}p_{3} \cdots p_{n} + p_{1}p_{3} \cdots p_{n} + \ldots + p_{1}p_{2} \cdots p_{n-1}}{p_{1}p_{2} \cdots p_{n}}$, where $p_{j}$ is the $j$th prime number.
This can be brought to the form $\sum_{i=1}^{n}\frac{1}{p_{i}} = \frac{\alpha \boldsymbol{P} + \beta}{\boldsymbol{P}}$, where $\boldsymbol{P} = p_{1}p_{2} \cdots p_{n}$ and $\alpha, \beta$ are integers and $0< \beta < \boldsymbol{P}$.
Is there any way to put limits on $\beta$ other than the one listed above? I'm trying to see how close to an integer the sum can be for a given $n$. After plugging a few values into a spreadsheet, it appears that $ln(\beta) = \mathcal{O}\left(n\right)$, but I'm wondering if there is a way to show this or to put limits on $\beta$ (the lower limit for $\beta$ is what I'm interested in).
UPDATE
The equation above implies that $\prod_{i=1}^{n}p_{i} \sum_{i=1}^{n}\frac{1}{p_{i}} \equiv \beta \left( mod \prod_{i=1}^{n}p_{i} \right)$.
Since $ln(\beta) = \mathcal{O}\left(n\right)$, $\beta = \mathcal{O}\left(e^{n}\right)$ and the congruence above becomes
$$\prod_{i=1}^{n}p_{i} \sum_{i=1}^{n}\frac{1}{p_{i}} \equiv \mathcal{O}\left(e^{n}\right) \left( mod \prod_{i=1}^{n}p_{i} \right)$$
where the congruence is taken to be an "order of magnitude congruence" or "approximate congruence" (I'm not aware of an existing notation for such a thing). The curious thing is that the product of $n$ primes times the inverse sum of those same primes is congruent to approximately $e^{n}$ mod the product, so there must be tighter limits than $0< \beta < \prod_{i=1}^{n}p_{i}$ (for example, $\beta \neq 3$ and $\beta \neq \frac{1}{3}\prod_{i=1}^{n}p_{i}$ for large $n$).
This is not an answer but here is a heuristic that $\beta$ should be large: $\beta(n)$ is minimized for $n$ when $$\lfloor \sum_{i = 1}^n p_i^{-1}\rfloor - \lfloor \sum_{i = 1}^{n-1} p_i^{-1}\rfloor = 1$$ But $$\sum_{p \leq n}^n p^{-1} = \log\log(n) + O(1)$$ and $$\prod_{p \leq n}p \approx e^{n\log n}$$ for large $n$. So a mimimized $\beta(n)$ is approximatley going to be an integer in the interval $$[1, \frac{e^{e^{e^{\frac{n}{\log n}}}e^{\frac{n}{\log n}}}}{n \log n}]$$ And primes are sufficiently random that my guess is the $\beta(n)$ is going to be more or less sampled uniformly from this interval.