So I have this little problem I want to solve that says the following:
For every number $\alpha \in \mathbb{R} $, analyze the following sequence $$ \{ x_n \} = \frac{1}{n^2} \sum_{k=1}^{n} \left[ k \alpha \right] ,$$ where $ \left[ x \right] $ is the floor function. Calculate for which values of $ \alpha $ does $ x_n $ converges, or if it diverges. If it converges, calculate the limit of the sequence.
My first thought was to prove that this sequence converges, or diverges, for every possible value of $ \alpha $, that is, when $ \alpha \in \mathbb{Z} ,$ $ \alpha \in \mathbb{Q} ,$ and when $ \alpha \in \mathbb{I} ,$ in other words, $ \alpha $ is irrational.
So when $ \alpha \in \mathbb{Z} $ I got the following: If $ \alpha \in \mathbb{Z} $ then $ \left[ \alpha \right] = \alpha ,$ therefore $$ \{ x_n \} = \frac{1}{n^2} \sum_{k=1}^{n} k \alpha = \frac{\alpha}{n^2} \sum_{k=1}^{n} k = \frac{\alpha}{n^2} \frac{n(n+1)}{2} = \frac{\alpha (n+1)}{2n} ,$$ thus, $ \{ x_n \} $ converges to $$ \{ x_n \} = \frac{\alpha (n+1)}{2n} ;$$ since it converges, then we can calculate the limit as follows $$ \lim_{n \to \infty} \{ x_n \} = \lim_{n \to \infty} \frac{\alpha (n+1)}{2n} = \frac{\alpha}{2} \lim_{n \to \infty} \frac{n+1}{n} = \frac{\alpha}{2} \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{1} = \frac{\alpha}{2} \left[ \lim_{n \to \infty} 1 + \lim_{n \to \infty} \frac{1}{n} \right] = \frac{\alpha}{2} .$$
So, first of all, is this proof correct? If so, how can I proceed to prove the sequence when $ \alpha \in \mathbb{Q} $ or when it's irrational? Should I prove the floor function for rational and irrational numbers and then apply this to the sequence?
For $\alpha \in \mathbb{R}^+$ we have $\lfloor k \alpha\rfloor \leqslant k \alpha < \lfloor k \alpha \rfloor +1 \implies k\alpha -1 < \lfloor k \alpha \rfloor \leqslant k \alpha.$
Hence,
$$\alpha\frac{n(n+1)}{2n^2} - \frac{n}{n^2} =\frac{ 1}{n^2}\sum_{k = 1}^n (\alpha k - 1) < \frac{1}{n^2}\sum_{k = 1}^n \lfloor k \alpha \rfloor \leqslant \frac{ 1}{n^2}\sum_{k = 1}^n \alpha k = \alpha\frac{n(n+1)}{2n^2}.$$
Apply the squeeze theorem to evaluate the limit. The cases where $\alpha \leqslant 0 $ is similar