Consider the arithmetic progression with fixed $ n$, $ 1 + n \cdot k \forall k \in \mathbb N$ let $ p_{n_j} \in \mathbb P$ and $ p_{n_j} = 1 + n \cdot j$ (by Dirichlet's theorem on arithmetic progressions) $ \frac {1}{1 \cdot (1 + j) \cdot (1 + 2j) \cdots(1 + (n - 1)j)}$ $ \frac {1 + j}{1 \cdot (1 + j) \cdot (1 + 2j) \cdots(1 + (n - 1)j)}$ $ \frac {1 + 2j}{1 \cdot (1 + j) \cdot (1 + 2j) \cdots(1 + (n - 1)j)}$ $ \cdots$ $ \frac {1 + (n - 1)j}{1 \cdot (1 + j) \cdot (1 + 2j) \cdots(1 + (n - 1)j)}$ is maximal arithmetic progression of length $ n$ in $ S$. For Example: $ n=1996$ $ p_{1996_1}=1+1 \cdot 1996=1997$ is prime then: $ \frac{1}{1 \cdot 2 \cdot 3 \cdots 1996}$ $ \frac{2}{1 \cdot 2 \cdot 3 \cdots 1996}$ $ \frac{3}{1 \cdot 2 \cdot 3 \cdots 1996}$ $ \cdots$ $ \frac{1996}{1 \cdot 2 \cdot 3 \cdots 1996}$ is a maximal arithmetic progression of length $ 1996$ in $ S$ Another example: $ n=1997$ Pick $ p=1+1997 \cdot 4$ prime