A sum on Beatty's theorem

Let $\alpha, \beta$ be positive irrational numbers such that $\displaystyle \frac{1}{\alpha} + \frac{1}{\beta}=1$ . Evaluate the (pseudo) sum:

$\sum_{n=1}^{\infty} \left(\frac{1}{\lfloor n\alpha\rfloor^2}+\frac{1}{\lfloor n\beta\rfloor^2}\right)$

Solution

We are using Beatty’s theorem . In brief, it states that for positive irrational numbers $\alpha, \beta$ with $\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}=1$ the sequences $\lfloor \alpha\rfloor, \lfloor 2\alpha\rfloor, \lfloor 3\alpha\rfloor, \dots$ and $\lfloor \beta\rfloor, \lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor, \dots$ are complementary. (i.e. disjoint and their union is $\mathbb{N}$ ). Thus our sum is nothing else than $\displaystyle \zeta(2)=\frac{\pi^2}{6}$ .