Given a natural number $n>1$, add up all the fractions $\dfrac{1}{pq}$ , where $p$ and $q$ are relatively prime,
$0< p < q \le n$, and $p+q>n$. Prove that the result is always $\dfrac{1}{2}$.
What I Know:
1. The proof itself was done using induction.
What I Don't Know
1. I honestly have no idea how to write the proof to this problem.
Can somebody please help me?
Put $A_n=\{(p,q); 1\leq p<q\leq n, p\wedge q=1,p+q\geq n+1\}$
Then $A_{n+1}$ is different of $A_n$ in only two ways:
1) if $(p,q)$ is in $A_n$ and $p+q=n+1$, then $(p,q)$ is not in $A_{n+1}$ (let $B_n$ this set). Otherwise, $(p,q)$ is in $A_{n+1}$.
2) The elements $(p,n+1)$ with $p$ prime to $n+1$ are in $A_{n+1}$, but not in $A_n$.Otherwise, an element of $A_{n+1}$ is in $A_n$. Hence, the difference between the sum over $A_n$ and $A_{n+1}$ is $$-\sum_{(p,q)\in B_n}\frac{1}{pq}+\frac{1}{n+1}\sum_{p\wedge(n+1)=1}\frac{1}{p}$$
Now as $\frac{1}{p(n+1-p)}=\frac{1}{n+1}(\frac{1}{p}+\frac{1}{n+1-p})$, the sum over $B_n$, is equal to
$$\frac{1}{n+1}(\sum\frac{1}{p}+\sum\frac{1}{n+1-p})$$ where the sum is over the $p$ with $1\leq p<\frac{n+1}{2}$,$p$ prime to $n+1$ and we see that this is $\displaystyle \frac{1}{n+1}\sum_{p\wedge(n+1)=1}\frac{1}{p}$, and we are done.