How prove this irrational $x\in(0,1)$,then have $0
show that:
for any irrational $x\in(0,1)$,and positive integer $n$,there exsit positive integer $p_{1},p_{2},\cdots,p_{n}$ where
$$p_{1}<p_{2}<\cdots<p_{n}$$
such
$$0<x-\sum_{i=1}^{n}\dfrac{1}{p_{i}}<\dfrac{1}{n!(n!+1)}$$
This problem is from Mathematical contest in jiangxi province at last problem.
I think this reslut  maybe involves
Irrational Approximation?
Note $$\dfrac{1}{n!(n!+1)}=\dfrac{1}{n!}-\dfrac{1}{n!+1}$$
Thank you  for you help

The greedy algorithm works. We let $x_0 = x$, and for all $n$ define
$$p_{n+1} = \left\lfloor \frac{1}{x_n}\right\rfloor + 1; \quad x_{n+1} = x_n - \frac{1}{p_{n+1}}.$$
It is easily verified that
$$0 < x_1 = x - \frac{1}{p_1} < \frac{1}{2} = \frac{1}{1!(1!+1)},$$
and since all $x_n$ are irrational, we have $p_{n+1}-1 < \frac{1}{x_n} < p_{n+1}$, whence
$$0 < x_n - \frac{1}{p_{n+1}} < \frac{1}{p_{n+1}-1} - \frac{1}{p_{n+1}},$$
and by induction $p_{n+1} > n!(n!+1)$, which yields
$$\begin{align}
\frac{1}{p_{n+1}-1} - \frac{1}{p_{n+1}} &< \frac{1}{n!(n!+1)} - \frac{1}{n!(n!+1)+1}\\
&= \frac{1}{n!(n!+1)(n!(n!+1)+1)}\\
&\leqslant \frac{1}{(n+1)!((n+1)!+1)}.
\end{align}$$
