TITLE: How to prove that $\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}-\frac{1}{x_1x_2...x_n}\in \mathbb{N}\cup \{0\}$
QUESTION [7 upvotes]: Question:

Show that for every natural number $n$ there exist $n$ natural numbers $ x_1 < x_2 < ... < x_n ,$ such that 
  $$
\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}-\frac{1}{x_1x_2...x_n}
\in \mathbb{N}\cup \{0\}.
$$

My idea:
For $n=1$ we can take any natural $x_1\geq 1$, since $\dfrac {1}{x_1} - \dfrac {1}{x_1} = 0$.
For $n=2$ we can take $x_1=1$ and any natural $x_2>1$, since $\dfrac {1}{1} + \dfrac {1}{x_2} - \dfrac {1}{1\cdot x_2} = 1$. 
For $n=3$, we can take $x_1=2,\,x_2=3$ and $x_3=5$.
But for $n>3$, I can't find any example, Now  I have read this links,But I can't understand
can you someone have easy example?
Thank you for you help.

REPLY [4 votes]: The sequence shown by Gerry Myerson, ie. $a_1=2$, $a_{k+1}=a_1\cdots a_k+1$ satisfies
$$
\frac1{a_1}+\dots+\frac1{a_n} = 1 - \frac1{a_1\cdots a_n}.
$$
(Induction works due to $\frac1{a_1\cdots a_n} - \frac1{a_1\cdots a_{n+1}} = \frac{a_{n+1}-1}{a_1\cdots a_{n+1}}=\frac1{a_{n+1}}$.)
Based on this sequence, for $n\ge3$ the numbers
$$ (x_1,x_2,\ldots,x_n) = (a_1,\ldots,a_{n-1},a_n-2) $$
solves the problem:
$$
\frac1{x_1}+\dots+\frac1{x_{n-1}}+\frac1{x_n} = 
\left(\frac1{a_1}+\dots+\frac1{a_{n-1}}\right)+\frac1{a_n-2} = \\
= \left(1-\frac1{a_1\cdots a_{n-1}}\right)+\frac1{a_1\cdots a_{n-1}-1} =
1+\frac1{a_1\cdots a_{n-1}(a_1\cdots a_{n-1}-1)} = \\
= 1+\frac1{a_1\cdots a_{n-1}\cdot (a_n-2)}.
= 1+\frac1{x_1\cdots x_{n-1}\cdot x_n}.
$$