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: 
\begin{align}3^{2n+1}+2^{n+2}=3\cdot 9^n + 4\cdot 2^n&=7\cdot 9^n -4(9^n-2^n)\\
&=7\cdot 9^n-4(9-2)(9^{n-1}+9^{n-2}\cdot2+\cdots+2^{n-1})\\
&=7[ 9^n-4(9^{n-1}+9^{n-2}\cdot 2\cdots+2^{n-1})]\end{align}

Which is divisible by $7$

Here is a solution based on congruences,

$$3\cdot 9^n +4\cdot 2^n \equiv 3\cdot 2^n +4\cdot 2^n \equiv 7\cdot 2^n \equiv 0\text{(mod 7)}$$
We have

$$3^{2n+1}+2^{n+2}=3\times\color{red}9^n+4\times2^n\equiv3\times\color{red}2^n+4\times2^n=7\times2^n\equiv0\mod7$$