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 $2 \equiv 3^2 \pmod 7$, therefore
$$
3^{2n+1} + 2^{n+2} \equiv 3^{2n+1} + 3^{2n+4} \equiv 3^{2n+1} \cdot 28 \pmod 7,
$$
and that is of course divisible by $7$.
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 $7$.