Let $a\in \mathbb{R}$. Show that $\frac{\lfloor na \rfloor}{n} \to a$ for $n\to \infty$

Question

I am completely stumped when it comes to that question. Any impulses?

Hint: $na-1\leq \lfloor na \rfloor \leq na$ – Jakobian May 31 '19 at 14:28 — Jakobian, May 31 '19 at 14:28
@Analysis Use squeezing theorem. – FreeMind May 31 '19 at 14:33 — FreeMind, May 31 '19 at 14:33
Possible duplicate of Solving a sequence limit with floor – Martin R May 31 '19 at 15:34 — Martin R, May 31 '19 at 15:34

score 4 · Answer 1 · answered May 31 '19 at 14:30

4

By squeezing

$$\frac{na}{n}\le\frac{\lfloor na\rfloor}{n} <\frac{na+1}{n}.$$

If you prefer,

$$a\le\frac{\lfloor na\rfloor}{n} <a+\frac{1}{n}.$$

answered May 31 '19 at 14:30

score 0 · Answer 2 · edited May 31 '19 at 14:31

0

HINT: Notice that $0\le na-\lfloor na \rfloor \lt 1$ for all $n\in\mathbb Z$.

edited May 31 '19 at 14:31

answered May 31 '19 at 14:28

Franklin Pezzuti Dyer

39,754
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score 0 · Answer 3 · answered May 31 '19 at 15:12

0

For each integer $n$, there are $0<\epsilon_n<1$ such that $\lfloor na \rfloor=na+\epsilon_n$, so $\frac{\lfloor na \rfloor}{n}=a+\frac{\epsilon_n}{n}\to a$ as $n\to \infty.$

answered May 31 '19 at 15:12

Matematleta

29,139

Let $a\in \mathbb{R}$. Show that $\frac{\lfloor na \rfloor}{n} \to a$ for $n\to \infty$

3 Answers3