<s>[INST] 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.
 [/INST] If $p+q=n+1$ and $(p,n+1)=1$, then it follows from Euclidean Algorithm that $(p,q)=1$. Now, we are to prove:
$$\frac{1}{n+1} \cdot \sum_{1 \leqslant p<n+1}{\frac{1}{p}}=\sum_{1 \leqslant p<n+1}{\frac{1}{pq}}=\frac{1}{2}\sum{\frac{1}{p(n+1-p)}}$$
This can be seen easily by writing: $$\frac{1}{p(n+1-p)}=\frac{1}{n+1}\bigg(\frac{1}{p}+\frac{1}{(n+1-p)}\bigg)$$
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