Can $\sum_{k\in M}\frac{1}{k}$ be a large integer?

Solution 1:

Yes, and more is true: given any positive rational number $n$, there exists a finite set $M\subset\Bbb N$ such that $\sum_{k\in M} \frac1k = n$.

Perhaps the most straightforward proof is this: since the harmonic series diverges, there exists a unique $m\in\Bbb N$ such that $$ \sum_{k=1}^m \frac1k \le n < \sum_{k=1}^{m+1} \frac1k. $$ (If $n$ is small then $m$ might equal $0$.) Write $r=n-\sum_{k=1}^m \frac1k$, which is a rational number less than $\frac1{m+1}$. Then use the greedy algorithm to write $r$ as an Egyptian fraction $\sum_{k\in M_r} \frac1k$. By size considerations, every element of $M_r$ exceeds $m$, and so $M=\{1,\dots,m\} \cup M_r$ has the property that $\sum_{k\in M} \frac1k = n$.

Similar constructions can yield representations of $n$ with particular constraints; for example, one can choose any $j\in\Bbb N$ and force all the elements of $M$ to exceed $j$.

Solution 2:

Clearly for any $n_0$ there exists a (not necessarily integer) $N$ and a set $S$ such that $n \le N < n + 1 $ by the divergence of the harmonic series with: $$ N = \sum_{k \in S} \frac1k $$

Let $ q = n + 1 - N $. It is well known that there exists a finite decomposition of $q$ into unit fractions with unique denominators (call the multiset of such $Q$). Using the identity:

$$ \frac{1}{a} = \frac{1}{a+1} + \frac{1}{a(a+1)}$$

we can make the denominators in $Q$ larger than the largest element of $S$. (We have to be careful how we do this, but clearly it is possible.) Since $Q$ is finite, this process terminates. Now $Q \cup S $ is the desired set with $\sum_{k \in Q \cup S} \frac1k = n + 1$, and $Q \cap S = \emptyset$.

Solution 3:

The Erdős–Graham conjecture in combinatorial number theory states that, if the unit fractions are partitioned into finitely many subsets, then one of the subsets has a finite subset of itself whose reciprocals sum to one.

That would imply that we can write every positive integer as sum of unit fractions: We show that we can construct a sequence $(A_n)_{n\in\mathbb N}$ of disjoint finite subsets of $\mathbb N\setminus \{1\}$ such that $\sum_{i\in A_n} \frac{1}{i}=1$ for all $n\in\mathbb N$. We construct the sequence by induction: Let $n\ge 1$ be given and $A_1 ,\dots A_n$ be already defined. We split each of those $A_i$ in two non-empty set and define $$B:=\mathbb N \setminus \left( \{1\} \cup \bigcup_{i=1}^n A_i \right).$$ Then those $2n+1$ sets are a partition of $\mathbb N\setminus\{1\}$. By the stated conjecture $B$ contains a finite subset $B'$ such that $\sum_{i\in B'} \frac{1}{i}=1$. We define $A_{n+1}:=B'$.

In order to write a given positive integer $n$ as sum of unit fractions we simply define $$S:= \bigcup_{i=1}^n A_i $$ and have $$\sum_{i\in S} \frac{1}{i}=n.$$