Proof that a sequence does not converge

Question

So here is the question:

Prove that the following sequence does not converge: $a_n= \frac{2n}{5}- \lfloor \frac{2n}{5} \rfloor.$ Note that $\lfloor x \rfloor$ = max {$z: z$ is an integer and $z \leq x$}.

What it means in a practical sense is that if x is positive and has a decimal expansion, the floor will be the unique integer obtained by taking the decimal expansion of x and removing all terms to the right of the decimal.

My proof so far: We will be using proof by contradiction. Assume that $a_n= \frac{2n}{5}- \lfloor \frac{2n}{5} \rfloor$ converges to some $L \in reals$. Let $\epsilon=...$

This is what I have so far. Let me know if you have any suggestions! Thank you in advance!

Answer 1

Like @dxiv said: $n=5k\implies a_n=0$

Now notice that $n=5k+1$ leads to the following sequence: $a_n=\frac{10k}5+\frac 25-\lfloor \frac{10k}5+\frac25\rfloor=2k+\frac 25-\lfloor 2k+\frac25\rfloor=2k+\frac 25-2k=\frac25$

So we have one subsequence converging to $0$ and one to $\frac25$

Answer 2

In this case it's easiest to just calculate what the sequence is.

$a_{5k} = 0$

$a_{5k + 1} = \frac 25$

$a_{5k + 2} = \frac 45$

$a_{5k+3} = \frac 15$

$a_{5k + 4} = \frac 35$.

Such a sequence is obviously not Cauchy so it will not converge. (No periodic sequence can converge.)