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 $7$.
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 $k\_0=1, k\_1=5$