# Find a necessary and sufficient condition

Let $$(a_n)_{n=1}^\infty$$ be a real sequence. Find a necessary and sufficient condition for $$(a_n)$$ so $$(\lfloor a_n \rfloor)_{n=1}^\infty$$ converges to $$0$$.

Hi everyone. I am trying to brush up on calculus. I believe the necessary and sufficient condition would be $$\lim_{n\to\infty}{a_n} \in [0,1).$$ It's clear for me why this is true (or could be) but I need to prove this using the definition of a limit. I've tried some tricks with the triangle inequality and the properties of the floor function, but for some reason I can't prove this properly.

I would love to hear your thoughts.

Your condition is sufficient, but not necessary. Take, for instance$$a_n=\begin{cases}\frac12&\text{ if n is odd}\\0&\text{ otherwise.}\end{cases}$$Then $$\lim_{n\to\infty}a_n$$ doesn't exist, but $$\lim_{n\to\infty}\lfloor an\rfloor=0$$.

A condition which is both necessary and sufficient is that $$n\gg1\implies a_n\in[0,1)$$. Can you prove it?

$$(\lfloor a_n \rfloor)$$ is a stationnary sequence.

If $$\lfloor a_n\rfloor$$ goes to zero, then for large enough $$n$$,

$$-\frac 12<\lfloor a_n\rfloor <\frac 12$$

thus $$\lfloor a_n\rfloor=0$$

and

$$0\le a_n<1$$

• This is valid only if you interpret $\lfloor a_n\rfloor$ as rounding to nearest, but normally it is the floor function. – gammatester Nov 25 '18 at 17:26
• Thank you very much Hamam! – Noy Perel Nov 25 '18 at 17:32