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
 A: 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
$$\begin{align}
\frac{1}{p_{n+1}-1} - \frac{1}{p_{n+1}} &< \frac{1}{n!(n!+1)} - \frac{1}{n!(n!+1)+1}\\
&= \frac{1}{n!(n!+1)(n!(n!+1)+1)}\\
&\leqslant \frac{1}{(n+1)!((n+1)!+1)}.
\end{align}$$
A: Proceed by induction. For $n=0$ the result is trivial. Suppose we have
$$0<x-\sum_{i=1}^n \frac{1}{p_i}<\frac{1}{n!(n!+1)}$$
and let 
$$p_{n+1}=\left\lceil \frac{1}{x-\sum_{i=1}^n \frac{1}{p_i}}\right\rceil$$
then we have
$$\frac{1}{p_{n+1}}<x-\sum_{i=1}^n \frac{1}{p_i}<\frac{1}{p_{n+1}-1}$$
so
$$0<x-\sum_{i=1}^{n+1} \frac{1}{p_i}<\frac{1}{p_{n+1}-1}-\frac{1}{p_{n+1}}=\frac{1}{p_{n+1}(p_{n+1}-1)}$$
Since $\frac{1}{p_{n+1}}<\frac{1}{n!(n!+1)}$ we have $p_{n+1}\ge n!(n!+1)+1$ so
$$\frac{1}{p_{n+1}(p_{n+1}-1)}\le \frac{1}{(n!(n!+1)+1)n!(n!+1)}\le \frac{1}{((n+1)!+1)(n+1)!}$$
A: Pick $p_1,\ldots ,p_n$ greedily, i.e. such that
For $1\le k\le n$ define 
$$\Phi(k)\quad\equiv \quad 1<p_1<p_2<\ldots <p_k\ \land\ 0<x-\sum_{i=1}^k\frac 1{p_i}<\frac1{p_k^2-p_k}\le \frac1{k!(k!+1)}.$$
For a given irrational $x\in(0,1)$, we shall define a sequence $(p_i)_{i=1}^\infty$ recursively in such a manner that $\Phi(k)$ holds for all $k\in\mathbb N$. Note that we never have to worry about equality in the second condition because $x$ is irrational.
The first $n$ terms of this sequence obviously solve the problem.
To achieve $\Phi(1)$, note that the conditions merely require $p_1\ge 2$ and $\frac 1{p_1}<x<\frac 1{p_1-1}$. Thus all is well if we pick $p_1=\lceil \frac1x\rceil$. 
Assume we have achieved $\Phi(k)$ for some $1\le k<n$.
Pick $$p_{k+1}=\left\lceil\frac1{x-\sum_{i=1}^k\frac 1{p_i}}\right\rceil.$$
From $\Phi(k)$ we conclude that $p_{k+1}>p_k(p_k-1)\ge p_k$. Thus we already have $$1<p_1<\ldots <p_{k+1}$$ and $$0<x-\sum_{i=1}^{k+1}\frac1{p_i}<\frac1{p_{k+1}-1}-\frac1{p_{k+1}}=\frac1{p_{k+1}^2-p_{k+1}}.$$
Remains to show that
$$\tag1 p_{k+1}^2-p_{k+1}\ge(k+1)!((k+1)!+1)$$
holds.
If $k=1$, this just means $p_2^2-p_2\ge 6$, which follows from $p_2\ge 1+p_2\ge 3$.
If $k=2$, we either have $p_1=2, p_2=3$ and then $p_3\ge7$ because $\frac12+\frac13+\frac16$ would be too big; or we have $p_1\ge 3$,  $p_2\ge p_1(p_1-1)\ge 6$, and again $p_3\ge 7$. At any rate, $p_3^2-p_3\ge 42=3!(3!+1)$.
If $k=3$ we verify directly that 
$$\tag2 k!(k!+1)\ge (k+1)^2.$$
If $k>3$, $$\begin{align}k!(k!+1)-1&\ge k(k-1)(k-2)\cdot 7-1\\
&=(k+1)^2+7(k-4)k^2+6k^2+11k+(k-2)\\
&\ge(k+1)^2 \end{align}$$
so that $(2)$ holds for all $k\ge 3$. 
Thus from $p_{k+1}>k!(k!+1)$ we have
$$p_{k+1}^2-p_{k+1}> k!(k!+1)\left(k!(k!+1)-1\right)\ge (k+1)^2k!(k!+1)>(k+1)!((k+1)!+1).$$
So finally we have shown $(1)$ for all $k\ge 1$. 
