For $N\in \mathbb{N}$, do there exist natural numbers $N<n_1<n_2<\cdots<n_k$ such that $\frac{1}{n_1}+\cdots+\frac{1}{n_k}=1$?

Solution 1:

You may start from: $$ \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1 \tag{1}$$ and use the identity: $$ \frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)} \tag{2}$$ to increase the number of terms in the LHS of $(1)$. Quoting Wikipedia, any fraction $\frac{x}{y}$ has an Egyptian fraction representation in which the maximum denominator is bounded by $O\left(\frac{y\log^2 y}{\log\log y}\right)$ and a representation with at most $O(\sqrt{\log y})$ terms.

Another possible approach is the following one: $$ \frac{1}{N}\prod_{k=N}^{N^2-1}\left(1+\frac{1}{k}\right)=1,\tag{3}$$ and if we expand the LHS of $(3)$ we get a sum of Egyptian fractions with a denominator $\geq N$.

We just need to remove duplicates: for such a task, we may use $(2)$.

Solution 2:

Jack D'Aurizio best me to it with a cleaner solution, but I'll post this anyway. We will use the result that,

For any rational $q \in \Bbb Q,$ there exists finitely many $n_1 < n_2 < \dots <n_k$ such that, $$ \sum_{i=1}^k \frac1{n_i} = q $$

There are many known algorithms of constructing such an expansion, but in particular there is a simple, greedy one that continuously adds the largest possible unit fraction such that the sum does not exceed $q.$ This is outlined on this page.

Then then let, $$ q = \sum_{n=1}^N \frac1n + 1 $$ Which will gives $n_1 < n_2 < \dots < n_k$ s.t. $$ \sum_{n=1}^N \frac1n + 1 = \sum_{n=1}^N \frac1n + \sum_{i=1}^k \frac1{n_i} $$ Which gives the desired result.