$\sum_{(p,q) \in {\mathbb{N}^*}^2 and p \land q =1} \frac{1}{p^2 q^2} = \frac{5}{2}$ proof?

Question

Can you give me a very precise demonstration of this result please because it's very difficult for me to understand the demonstration on the pic :(

$$ \sum_{(p,q) \in {\mathbb{N}^*}^2 \text{, } p \land q =1} \frac{1}{p^2 q^2} = \frac{5}{2} $$

Thank you ! Shadock

Answer 1

Let $A$ be the set of all coprimes pairs $(a,b)$ of natural numbers. There is a canonical bijection between $A\times\Bbb N$ and $\Bbb N\times\Bbb N$ given by $((a,b),g)\mapsto(ag,bg)$. This encodes the fact that after dividing a pair of naturals by their gcd one obtains a pair of coprime naturals, and one can go back by just multiplying by the gcd. Therefore, using the substitution $(x,y)=(ga,gb)$, we have

$$\zeta(2)^2=\sum_{(x,y)\in\Bbb N^2}\frac{1}{x^2y^2}=\sum_{\substack{(a,b)\in A \\ g\in\Bbb N}}\frac{1}{(ga)^2(gb)^2}=\sum_{g\in\Bbb N}\frac{1}{g^4}\sum_{(a,b)\in A}\frac{1}{a^2b^2}=\zeta(4)\sum_{(a,b)\in A}\frac{1}{a^2b^2}$$

and hence

$$\sum_{(a,b)\in A}\frac{1}{a^2b^2}=\frac{\zeta(2)^2}{\zeta(4)~}. $$

Answer 2

Ok, the solution itself is not easy to come by yourself, but it is not too hard to follow it.

The main points are:

• first line, you simply rewrite the square of a sum as the sum of all elements
• the real trick is that now that you're summing on $\mathbb{N}^2$, you can rewrite this, as long as you make sure that you still sum on $\mathbb{N}^2$. That's what the author is doing by summing on all the couple $(k,q',p')$ instead of $(q,p)$. You indeed have a bijection between the 2, so you are not modifying the overall sum by summing on that new tuple with that condition.
• the author then factors by $k$ the sum and renames the variables (which has no effect mathematically)
• the last lines are "usual" reseult on series which you seem to know about

If it is the operations on the series that you find hard to follow, you should try on small example to understand exactly what they mean

I hope this helps you a bit.

Answer 3

I'll assume that $p\land q=\gcd(p,q)$.

Since $\sum_{n\ge1}{1\over n^2}$ converges absolutely, we can treat this infinite series without serious consideration. We have: $$\begin{align} \left(\sum_{n\ge1}{1\over n^2}\right)^2 &=\left({1\over1^2}+{1\over2^2}+\cdots\right)\left({1\over1^2}+{1\over2^2}+\cdots\right) \\&=\sum_{m\ge1,n\ge1}{1\over m^2n^2} \\&=\sum_{m\ge1,n\ge1}{1\over (m\land n)^4}{1\over (m/(m\land n))^2(n/(m\land n))^2}. \end{align}$$ In the above equation, $m\land n$ can be any positive integer, and $(m/(m\land n),n/(m\land n))$ can be any pair of positive integers which are coprime. Actually,

choosing values of $(m/(m\land n),n/(m\land n))$ first and then choosing a value of $m\land n$

is same as

choosing values of $(m,n)$ at once,

since both determine each other uniquely. For example, (2,3) and 5 determines (10,15), and (10,15) determines (2,3) and 5.

Thus the above equation is: $$\begin{align} &=\sum_{a\ge1,b\ge1,a\land b=1 \atop c\ge1}{1\over c^4}{1\over a^2b^2} \\&=\sum_{c\ge1}{1\over c^4} \cdot \sum_{a\ge1,b\ge1,a\land b=1}{1\over a^2b^2}. \end{align}$$

We already have: $$\begin{gather} \sum{1\over n^2}={\pi^2\over6},\\ \sum{1\over n^4}={\pi^4\over90}, \end{gather} $$ (these are famous results. You can refer http://en.wikipedia.org/wiki/Basel_problem.)

so we get the result.