property of prime numbers

Question

I have the following similar sums: $$S_n = p_n - \sum_{i=1}^{n-1} \lfloor\frac{p_n}{p_i} \rfloor + \sum_{1\leq i_1 < i_2\leq n-1} \lfloor \frac{p_n}{p_{i_1}p_{i_2}}\rfloor + ... + (-1)^k \sum_{1\leq i_1 < ... < i_k\leq n-1}\lfloor \frac{p_n}{p_{i_1}...p_{i_k}}\rfloor + ...$$ and $$ T_n = p_n - \sum_{i=1}^{n-1} \frac{p_n}{p_i} + \sum_{1\leq i_1 < i_2\leq n-1} \frac{p_n}{p_{i_1}p_{i_2}} + ... + (-1)^k \sum_{1\leq i_1 < ... < i_k\leq n-1} \frac{p_n}{p_{i_1}...p_{i_k}} + ...$$ and $$Z_n = p_n - \sum_{i=1}^{n-1} \lceil\frac{p_n}{p_i} \rceil + \sum_{1\leq i_1 < i_2\leq n-1} \lceil \frac{p_n}{p_{i_1}p_{i_2}}\rceil + ... + (-1)^k \sum_{1\leq i_1 < ... < i_k\leq n-1}\lceil \frac{p_n}{p_{i_1}...p_{i_k}}\rceil + ...$$ where $p_i$ is the $i'th$ prime number. I think that $\lfloor \frac{p_n}{p_{i_1}...p_{i_k}}\rfloor = \lceil \frac{p_n}{p_{i_1}...p_{i_k}}\rceil - 1$, because $\lceil x \rceil - 1 = \lfloor x \rfloor$ if $x \in \mathbb{R}^*_+\setminus \mathbb{N} $, right? Therefore: $$ S_n = Z_n - \left( -\sum_{i=1}^{n-1} 1 + \sum_{1\leq i_1 < i_2\leq n-1} 1 + ... + (-1)^k \sum_{1\leq i_1 < ... < i_k\leq n-1}1 + ...\right)$$. I also think that $ \sum_{1\leq i_1 < ... < i_k\leq n-1}1 = C_{n-1}^k$ therefore it comes out that $S_n = Z_n - (-1) = Z_n + 1$. If this is correct, I think that the following property is true: $$ S_n > T_n > Z_n$$ but I do not know how to prove it ...