Splitting $\frac{1}{n}$ for $n\geq 2$ as a sum of $m\geq 2$ unit fractions (Various proofs)

Question

So the problem is to write $\frac{1}{n}=\sum_{1}^{m}\frac{1}{a_{k}}$ for $a_{k}\in \mathbb{N}$ (distinct if it is too easy).

The only proof I've seen is with $\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}$.

Do you know any other proofs or any other algorithms?

I am trying to play with Taylor expansion.

Thanks

Answer 1

Assuming you mean that $m$ is fixed, here is a particular case: the Erdos-Strauss conjecture says that:

$$\frac{4}{n}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

has a solution for each $n\geq 0$, with $x,y,z\in\mathbb{N}$. So for $m=3$, it is safe to say that this problem beyond the reach of current mathematics. A special case of this is known when $n\equiv 2\pmod 3$:

$$\frac{4}{n}=\frac{1}{n}+\frac{1}{(n-2)/3+1}+\frac{1}{n((n-2)/3+1},$$

but for $m\equiv 1\pmod{3}$, in particular $n=4k$ with $k\not\equiv 1\pmod{3}$, there is nothing known.

For general $m$, one can sometimes get lucky by employing the greedy algorithm for Egyptian fractions.

Answer 2

Let's look at the case $m=2$; we want all the ways of writing $1/n$ as a sum of two unit fractions. Note $${1\over n}-{1\over n+k}={k\over n(n+k)}$$ so we'll get a sum of two unit fractions if and only if $k$ divides $n(n+k)$, which is to say, if and only if $k$ divides $n^2$. Thus, for example, we get $${1\over6}={1\over7}+{1\over42}={1\over8}+{1\over24}={1\over9}+{1\over18}={1\over10}+{1\over15}={1\over12}+{1\over12}$$ although we're inclined to reject the last one as we want distinct denominators.

As a special case, if $n$ is prime, then the only $k$ dividing $n^2$ are $1,n,n^2$, and the only solution is $${1\over n}={1\over n+1}+{1\over n^2+n}$$ since we reject $1/n=1/(2n)+1/(2n)$.

One way to get solutions with $m>2$ is to apply this $m=2$ analysis to one term or the other arising in an $m=2$ solution. For example, there are 10 ways to express $1/24$ as a sum of two unit fractions (not counting $(1/48)+(1/48)$); combining these with $1/6=(1/8)+(1/24)$ gives 10 expressions for $1/6$ as a sum of three unit fractions.

Answer 3

For any integer $n>1$, let n = ab where $1 \le a < b \le n$.

Then $$\frac{1}{n} = \frac{1}{a(a+b)} + \frac{1}{b(a+b)}$$

It follows that $$\frac{1}{n} = \frac{1}{a(a+b)} + \frac{1}{b(a+2b)} + \frac{1}{(a+b)(a+2b)}$$

$$\frac{1}{n} = \frac{1}{a(a+b)} + \frac{1}{b(a+2b)} + \frac{1}{(a+b)(2a+3b)} + \frac{1}{(a+2b)(2a+3b)}$$

Clearly this pattern can be continued forever.