6. Find the sum of all fractions $\frac{1}{x y}$ such that: $\operatorname{gcd}(x, y)=1, x \leqslant n, y \leqslant n, x+y>n$.
6. Starting from special cases, give us insights.

For $n=2$, we have $x=1, y=2$ and $x=2, y=1$, the sum is $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 1}=1$.
For $n=3$, we need to consider the pairs $(1,3),(3,1),(2,3)$ and (3,2), the sum is $\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 1}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 2}=1$.
We conjecture that $S_{n}=\sum \frac{1}{x y}=1$, where the sum is over all pairs $(x, y)$ such that $x \leqslant n, y \leqslant n, x+y>n, \operatorname{gcd}(x, y)=1$. Suppose this is true for some $n$, how much does $S_{n+1}$ differ from $S_{n}$? When going from $n$ to $n+1$, the terms $\frac{1}{x y}$ in $S_{n}$ that satisfy $x+y>n+1$ are all in $S_{n+1}$, and the terms removed are those $\frac{1}{x y}$ where $x+y \leqslant n+1$. These are terms of the form $\frac{1}{x(n+1-x)}$. For each such removed fraction, there are two additional fractions $\frac{1}{x(n+1)}, \frac{1}{(n+1-x)(n+1)}$ in $S_{n+1}$. Since $x$ is coprime with $n+1$, $n+1-x$ is also coprime with $n+1$. Because $\frac{1}{x(n+1-x)} = \frac{1}{x(n+1)}+\frac{1}{(n+1-x)(n+1)}$, we have $S_{n}=S_{n+1}$.