Calculate the sum of fractionals 
Let $n \gt 1$ an integer. Calculate the sum: $$\sum_{1 \le p \lt q \le
 n} \frac 1 {pq} $$ where $p, q$ are co-prime such that $p + q > n$.


Calculating the sum for several small $n$ value I found out that the sum is always $\frac 1 2$.
Now, I'm trying to prove the sum is $\frac 1 2$ using induction by $n$. Suppose it's true for all values less or equals to $n$, trying to prove it for $n + 1$.
$$\sum_{1 \le p \lt q \le n +1,p+q>n+1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n} \frac 1 {pq} + \sum_{1 \le p \lt q = n +1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n} \frac 1 {pq} + \frac 1 {n+1} \sum_{1 \le p \lt n +1} \frac 1 {p}  \tag1$$
In the second sum, $p$ and $n+1$ are coprime.
$$\sum_{1 \le p \lt q \le n,p+q>n+1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n, p+ q>n} \frac 1 {pq} - \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} = \frac 1 2 - \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} \tag 2$$
From (1) and (2) I have to prove that $$\frac 1 {n+1} \sum_{1 \le p \lt n +1} \frac 1 {p} = \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} \tag3$$
where $p,q$ are co-prime and I'm stuck here.

Let $$
s_n=\sum_{1\le p<q\le n\\(p,q)=1}\frac1{pq}-\sum_{p+q\le n\\p<q,(p,q)=1}\frac1{pq}=a_n-b_n.
$$ Then we have that
$$
a_{n+1}-a_n=\sum_{1\le p<q=n+1\\(p,n+1)=1}\frac{1}p\cdot\frac1{n+1}=\frac1{n+1}\sum_{1\le p<n+1\\(p,n+1)=1}\frac1{p}
$$ and $$\begin{align*}
b_{n+1}-b_n&=\sum_{p+q=n+1\\p<q, (p,q)=1}\frac1{pq}
\\&=\frac1{n+1}\sum_{p+q=n+1\\p<q, (p,q)=1}\frac1 p+\frac 1 q
\\&=\frac1{2(n+1)}\sum_{p+q=n+1\\(p,q)=1}\frac1 p+\frac 1 q
\\&=\frac1{n+1}\sum_{p+q=n+1\\(p,q)=1}\frac1 p\tag{*}
\\&=\frac1{n+1}\sum_{1\le p<n+1\\(p,n+1)=1}\frac1 p\tag{**}.
\end{align*}$$
$(*)$ : $\displaystyle \sum_{p+q=n+1, (p,q)=1}\frac1 p=\sum_{p+q=n+1, (p,q)=1}\frac1 q$ by symmetry.
$(**)$ : $(p,q)=(p,p+q)=(p,n+1)=1$ by Euclidean algorithm.
This gives $s_{n+1}-s_n = (a_{n+1}-a_n)-(b_{n+1}-b_n)=0$, hence $s_n =s_2=\frac 12$ for all $n\ge 2$.
