
Math problem
Can anyone help me to solve this problem:
Consider $\displaystyle n$ as a natural number greater than $\displaystyle 1$. Prove that sum of all numbers of the form $\displaystyle \frac{1}{pq}$ with the condition $\displaystyle (p,q) = 1$ ($\displaystyle p$ and $\displaystyle q$ are relatively prime) and $\displaystyle 1 \leq p < q \leq n$ and $\displaystyle p + q > n$ is always equal to $\displaystyle \frac {1}{2}$

The problem doesn't seem to be right.
Take n=3, p=2, q=3. That's the only combination possible, but 1/pq= 1/6.

My bad. I mistyped the problem statement :(
It's $\displaystyle 1 \leq p < q \leq n$ instead of $\displaystyle 1 < p < q \leq n$
Now it's corrected.

Rule #10: Do not beg for answers. If your question is unanswered do not make a post begging for someone to reply back, it is totally useless. Other members can see the question.
But what should i do? I need the answer. I would wonder if anyone help me.
Thnx

Prime Induction?
I'll post more later, but fiddling with the problem, I believe it's equivalent to proving the following relation holds for odd prime p and natural k:
$\displaystyle \frac12\left(\sum_{i=1}^{p^k1}\frac1{i(p^ki)}\frac1p\sum_{i=1}^{p^{k1}1}\frac1{i(p^{k1}i)}\right)$ = $\displaystyle \frac1{p^k}\left(\sum_{i=1}^{p^k1}\frac1i\frac1p\sum_{i=1}^{p^{k1}1}\frac1i\right)$ (Wondering)
Prepare for another complicated problem...

Sorry for the delay...
Here's where I started. The problem states that for all $\displaystyle n$, $\displaystyle \sum_{S_n}\frac1{pq}=\frac12$, for $\displaystyle S_n=\{(p,q)\in\mathbb{N}^21\leq p <q\leq n, p+q>n, (p,q)=1\}$
Base case $\displaystyle n=2$: $\displaystyle S_2=\{(1,2)\}$, so $\displaystyle \Sigma=\frac12$
Look at $\displaystyle n=3,4,5,6...$
$\displaystyle S_3=\{(1,3)(2,3)\}$
$\displaystyle S_4=\{(1,4)(2,3)(3,4)\}$
$\displaystyle S_5=\{(1,5)(2,5)(3,4)(3,5)(4,5)\}$
$\displaystyle S_6=\{(1,6)(2,5)(3,4)(3,5)(4,5)(5,6)\}$
$\displaystyle S_7=\{(1,7)(2,7)(3,5)(3,7)(4,5)(4,7)(5,6)(5,7)(6,7 )\}$
$\displaystyle S_8=\{(1,8)(2,7)(3,7)(3,8)(4,5)(4,7)(5,6)(5,7)(5,8 )(6,7)(7,8)\}$
See if you can notice the pattern. First of all, most $\displaystyle (p,q)$ pairs stay in the set when moving from $\displaystyle n1$ to $\displaystyle n$. This is because if they are relatively prime, this has nothing to do with $\displaystyle n$. So, consider the pairs we gain advancing from $\displaystyle n1$ to $\displaystyle n$, and consider the pairs we lose.
Gain: From $\displaystyle n1$ to $\displaystyle n$, we gain all pairs $\displaystyle (p,n)=1$
Lose: From $\displaystyle n1$ to $\displaystyle n$, we lose all pairs $\displaystyle p+q=n$
Consider $\displaystyle L_n=lcm(1,2,3,...n)$. I observe exactly three cases...
Case 1: $\displaystyle L_n=L_{n1}$, n brings no new factors, so the only pair lost is $\displaystyle (1,n1)$ and the only pairs gained are $\displaystyle (1,n),(n1,n)$ Example $\displaystyle 5\to6$
Case 2: If $\displaystyle n$ is prime, $\displaystyle L_{n}=nL_{n1}$. Since n brings exactly one new factor, the pairs gained are $\displaystyle (1,n)(2,n)(3,n)...(n1,n)$ and the pairs lost are those of the form $\displaystyle (i,ni)$ for all $\displaystyle i=1,2,3,...n1$ Example $\displaystyle 6\to7$
Case 3: If n is a power of a prime $\displaystyle m^k$, $\displaystyle L_n=mL_{n1}$. So n brings the same gains/losses as case 2, excepting the pairs $\displaystyle (p,q)$ where the prime $\displaystyle mp,mq$ (sorry, poor choice of notation). Example $\displaystyle 7\to8$
The idea of this approach is that since most of the pairs stay the same, their "part" of the sum remains unchanged. So in jumping from n1 to n, the part of the sum subtracted from the lost pairs must balance perfectly the part of the sum added by the gained pairs. So for case 1 it is easy to verify the algebraic identity $\displaystyle \frac1{1*(n1)}=\frac1{1*n}+\frac1{(n1)*n}$
For case 2, we have the more complicated $\displaystyle \frac12\sum_{i=1}^{n1}\frac1{i(ni)}=\frac1n\sum_{i=1}^{n1}\frac1i$ for n prime
Case 3 is the more complicated formula from my last post. It can easily be seen that case 2 is a special case of 3 for k=1. So if we can prove case 3 the problem is done. However, it may be easier to tackle 2 first to gain enough insight to do 3.

On second thought...
Actually, turns out $\displaystyle \frac12\sum_{i=1}^{n1}\frac1{i(ni)}=\frac1n\sum_{i=1}^{n1}\frac1i$ is true for all n, not just primes, which makes the remainder of the proof pretty simple. I'll leave you the pleasure of proving the summation.