Question: Let $$S=\left\{\frac{1}{n}-\frac{1}{m}: m,n∈\mathbb{N}\right\}$$

Find the Supremum of this set.
I get the feeling that the answer is $1$ as if you let $n=1$ and $m$ be infinitely large then its $1-\text{something infinitely small}$, but I'm not sure how to prove this?

Answer: 
Note that a sequence with form $u\_n=a\cdot 9^n+b\cdot 2^n$ satisfies a linear recurrence with auxiliary equation $$(x-9)(x-2)=0$$ so that $$u\_{n+1}=11u\_n-18u\_{n-1}$$

Hence if $7|u\_n \& 7|u\_{n-1}$ then $7|u\_{n+1}$. We have $u\_0=7, u\_1=35$ both divisible by $7$, so with a base case established we are done by induction.

If $u\_n=7k\_n$, then $k\_n$ satisfies the same recurrence with $k\_0=1, k\_1=5$
Just surprised that the very crude induction method with $u\_n=3\cdot 9^n+4\cdot 2^n$ hasn't been posted yet - it goes $$u\_{n+1} = 3\cdot 9^{n+1}+4\cdot 2^{n+1}=7\cdot3\cdot 9^n+2\cdot 3\cdot 9^n+2\cdot 4\cdot 2^n=7\cdot 3\cdot 9^n+2\cdot u\_n$$ which is divisible by $7$ if $u\_n$ is divisible by $7$. Base case is $u\_0=7$. It depends on the fact that $9-2$ is divisible by $7$.