Calculate the sum of fractionals

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 $n\ge 2$.