# 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 $$s_n=\sum_{1\le p Then we have that $$a_{n+1}-a_n=\sum_{1\le p and \begin{align*} b_{n+1}-b_n&=\sum_{p+q=n+1\\p $$(*)$$ : $$\displaystyle \sum_{p+q=n+1, (p,q)=1}\frac1 p=\sum_{p+q=n+1, (p,q)=1}\frac1 q$$ 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 $$n\ge 2$$.
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: $$\frac{1}{n+1} \cdot \sum_{1 \leqslant p
This can be seen easily by writing: $$\frac{1}{p(n+1-p)}=\frac{1}{n+1}\bigg(\frac{1}{p}+\frac{1}{(n+1-p)}\bigg)$$