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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.

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2 Answers 2

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$(\lfloor a_n \rfloor)$ 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

$$0\le a_n<1$$

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    $\begingroup$ This is valid only if you interpret $\lfloor a_n\rfloor$ as rounding to nearest, but normally it is the floor function. $\endgroup$ Commented Nov 25, 2018 at 17:26
  • $\begingroup$ Thank you very much Hamam! $\endgroup$
    – Noy
    Commented Nov 25, 2018 at 17:32
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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?

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