What is the set of accumulation points of $a_{n} = \frac{2n^2}{7} - \left\lfloor { \frac{2n^2}{7}}\right\rfloor$?

Question

What is the set of accumulation points of $a_{n} = \frac{2n^2}{7} - \left\lfloor { \frac{2n^2}{7}}\right\rfloor$?

I have calculated them by writing the first 12 terms of the sequence and found them to be $\{0, 1/7, 2/7, 4/7\}$ am I correct?

Is their another way of solving this type of questions without calculating the terms?

Answer 1

You should prove that each term of the sequence is one of the four numbers that you got and that each of these numbers appears infinititely often. It will follow from that that the set of accumulation points is $\left\{0,\frac17,\frac27,\frac47\right\}$.