Suppose that $A$ is a set of cardinality, say $n$. Let $$H_n = 1+1/2+\dots +1/n.$$ If $\phi:A\to \{1,2,\dots,n\}$ is one-one, then obviously $$\sum_{a\in A}\phi(a)^{-1} = H_n$$ I'm wondering if the converse of the above statement is true, that is, if $\phi:A\to \{1,2,\dots,n\}$ is such that $$\sum_{a\in A}\phi(a)^{-1} = H_n,$$ then should $\phi$ be necessarily one-one?

It looks to be true at least for small integers say $n=2,3,4$. I have feel that this maybe true. However, I couldn't show this in general. Any hint towards its solution is greatly appreciated.

up vote 1 down vote accepted

If some $\frac{1}{k}$ terms “disappear” in the sum, the same amount of them should be “overlapping” the terms that are present in the sum. We should show that there’s no chance for it to occur. Unfortunately, that is possible:

$$1+\frac{1}{2}+...+\frac{1}{20}=1+ \frac{1}{2}+ \frac{1}{ 3}+ \frac{1}{5}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{6}+...+ \frac{1}{9}+ \frac{1}{11}+...+ \frac{1}{14}+ \frac{1}{16}+...+ \frac{1}{20}+ \frac{1}{20}$$

by the equality that

$$\frac{1}{4}+ \frac{1}{10}+ \frac{1}{15}= \frac{1}{5}+ \frac{1}{6}+ \frac{1}{20}$$

See the Wikipedia article Egyptian Fraction for more results about the sum of unit fractions.

  • Thank you. Can you give some intuition on how you guessed it? – Kumara Dec 7 at 7:45
  • You can read the Wikipedia article I just added to my answer. I took an example directly from there. – William Sun Dec 7 at 7:49

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