In this exercise, I have the sequence $a_n = \lfloor \frac{n}{6}\rfloor$ and I have to prove that it's increasing but not strictly increasing.
I actually got stuck because I haven't been taught the difference between increasing and strictly increasing. I tried to search around, but tell me if I'm right: if $a_{n+1} \geq a_n$ then it's increasing. If $a_{n+1} > a_n$ then it's strictly increasing.
Right?
Anyway, about the proof I thought this:
$$a_{n+1} > a_n \implies \lfloor \frac{n+1}{6}\rfloor > \lfloor \frac{n}{6}\rfloor$$ Hence
$$\lfloor \frac{n}{6}\rfloor + \lfloor\frac{1}{6}\rfloor > \lfloor\frac{n}{6}\rfloor$$
But $\lfloor \frac{1}{6}\rfloor = 0$, hence $a_{n+1}$ is actually equals to $a_n$, hence we need $a_{n+1} \geq a_n$, so the sequence is not strictly increasing, but it's increasing.
Can you please tell me if this could work, or if I need more things / details?
Thank you!
You seem to think that if $a_{n+1}\geqslant a_n$ then $(a_n)_{n\in\Bbb N}$ is increasing and if $a_{n+1}>a_n$ then $(a_n)_{n\in\Bbb N}$ is strictly increasing. Actually, this simply doesn't make sense, since you don't say what $n$ is. Asserting that $(a_n)_{n\in\Bbb N}$ is increasing means that for each natural $n$ you have $a_{n+1}\geqslant a_n$ , and asserting that $(a_n)_{n\in\Bbb N}$ is strictly increasing means that for each natural $n$ you have $a_{n+1}>a_n$.
Your sequence is increasing because, for each natural $n$,\begin{align}\left\lfloor\frac{n+1}6\right\rfloor&=\left\lfloor\frac n6+\frac16\right\rfloor\\&\geqslant\left\lfloor\frac n6\right\rfloor.\end{align}And it is not strictly increasing because, for instance, $\left\lfloor\frac26\right\rfloor=\left\lfloor\frac16\right\rfloor$.
On the other hand, it is false in general that $\left\lfloor\frac n6+\frac16\right\rfloor=\left\lfloor\frac n6\right\rfloor+\left\lfloor\frac16\right\rfloor$.