Japan Mathematical Olympiad

We will show that there exists no positive integer $n$ for which the statement of the problem is satisfied. Let us denote by $A_n$ the set of all rational numbers $r$ for which there exist an integer $b$ and nonzero integers $a_1,\dots ,a_n$ such that $r=b+{\sum }_{k=1}^n\frac{1}{a_k}$. In order to establish our claim, it suffices to show that for any positive integer $n$ the set $A_n$ does not exhaust the set of all rational numbers. We will show more: in fact there exists a pair $p_n,q_n$ of rational numbers with $p_n<q_n$ such that there exists no member of the set $A_n$ which lies strictly in between $p_n$ and $q_n$. We will show the existence of such a pair recursively.

For $n=1$ it suffices to take $p_1=\frac{1}{3}$ and $q_1=\frac{1}{2}$. Now suppose, we were able to choose for some $n$ a pair $p_n,q_n$ such that $p_n<q_n$ for which there is no element from the set $A_n$ lying strictly between $p_n$ and $q_n$. Set $d=q_n-p_n$, and define $p_{n+1}=p_n+\frac{d}{3}$ and $q_{n+1}^{\prime }=q_n-\frac{d}{3}$. Suppose that there exist an integer $b$ and nonzero integers $a_1,\dots ,a_{n+1}$ for which $$p_{n+1}<b+\sum _{k=1}^{n+1}\frac{1}{a_k}<q_{n+1}^{\prime }$$ holds.

By the induction hypothesis, we have for each $l\in \{1,\dots ,n+1\}$, either $$p_n\ge b+\sum _{k=1,k\ne l}^{n+1}\frac{1}{a_k}$$ or $$b+\sum _{k=1,k\ne l}^{n+1}\frac{1}{a_k}\ge q_n.$$

If the former holds, then we have $\frac{1}{a_l}<p_{n+1}-p_n=\frac{d}{3}$, while if the latter holds, then $\frac{1}{a_l}<q_{n+1}^{\prime }-q_n=-\frac{d}{3}$. In either case, we get $-\frac{3}{d}<a_l<\frac{3}{d}$, which implies that for each $l\in \{1,\dots ,n+1\}$, there are only finitely many possible integral values that $a_l$ can take. Furthermore if we fix $(a_1,\dots ,a_{n+1})$, the number of integers $b$ that satisfy the property $p_{n+1}<b+{\sum }_{k=1}^{n+1}\frac{1}{a_k}<q_{n+1}^{\prime }$ is also finite.

Consequently, we can conclude that there exist at most a finite number of elements of the set $A_{n+1}$ which lie in between $p_{n+1}$ and $q_{n+1}^{\prime }$. If we define $q_{n+1}$ to be equal to the minimum of such exceptional numbers of $A_{n+1}$ if such exists and equal to $q_{n+1}^{\prime }$ if there is no such exceptional element, then we can see that $p_{n+1}$ and $q_{n+1}$ will satisfy the statement of our assertion for $n+1$, and therefore, our induction process is complete and our assertion is proved.