It is well-known that if we have a fraction $\frac ab$, with $a,b\in\mathbb N$ and $a<b$, if we apply the greedy algorithm in order to express it as a sum of unit fractions, then we may get fractions with very large denominators. For instance, if we apply the algorithm to $\frac5{31}$, what we get is$$\frac17+\frac1{55}+\frac1{3\,979}+\frac1{23\,744\,683}+\frac1{1\,127\,619\,917\,796\,295}$$and if we apply it to $\frac{1\,197}{2\,273}$, then the decimal representation of the last denominator has $14\,583$ digits. All this suggests that the following statement is true:
If $R>0$, then there are natural numbers $a$ and $b$ such that $a<b$ and that, if we apply the greedy algorithm to express $\frac ab$ as the sum of unit fractions, then one of the unit fractions $\frac1D$ that we get in that expression will be such that $\frac Db>R$.
My guess is that either this statement has already been proved or that it has already been stated as a conjecture. It that's so, can someone please provide a reference?
Best Answer
There's a very simple answer to the original question. The greedy algorithm, applied to $a/b=2/(2n-1)$, gives $${2\over2n-1}={1\over n}+{1\over n(2n-1)}$$ so in the notation of the original question we have $${D\over b}={n(2n-1)\over2n-1}=n$$ which can be made arbitrarily large.