maybe this conjectures is also hold? for any irrational $x\in(0,1)$,and positive integer $n$,there exsit prime numbers  $p_{1},p_{2},\cdots,p_{n}$ where
$$p_{1}<p_{2}<\cdots<p_{n}$$
such
$$0<x-\sum_{i=1}^{n}\dfrac{1}{p_{i}}<\dfrac{1}{n!(n!+1)}$$
I have read this  problem has solve exist postive integer $p_{1},p_{2},\cdots,p_{n}$
I conjectures also hold exist prime sequence? I feel the three solutions for this situation can't explain，so  How to solve this case?Thanks
 A: False.  Take $n=3$ and $x=\frac{41}{42} - \epsilon$ where $\epsilon>0$ is a small irrational (to be made precise later).  Then the claim is that there exist three primes $p_1 < p_2 < p_3$ such that
$$\frac{40}{42}  < \frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} + \epsilon < \frac{41}{42}.$$
Since $\frac13 + \frac15 + \frac17 = \frac{71}{105}$ is too small to satisfy the left inequality (assuming $\epsilon < \frac{29}{105}$), we must have $p_1 = 2$.  Similarly, since $\frac12 + \frac15 + \frac17 = \frac{59}{70}$ is too small (assuming $\epsilon < \frac{23}{210}$), we must have $p_2 = 3$.
This leaves the following inequality for $p_3$:
$$\frac{5}{42} < \frac{1}{p_3} + \epsilon < \frac{1}{7}.$$
Clearly $p_3 \le 7$ is not possible without violating the right inequality, so $p_3 > 7$.  But for any $\epsilon < \frac{13}{462}$, the left inequality is equivalent to $p_3 < 11$, which means no such prime exists.
Thus a counterexample is obtained for any irrational $\epsilon < \frac{13}{462} \approx 0.02814$.  We may easily take, for instance, $\epsilon = 0.01e$.
