Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the sequence ${x_n}$ defined by $x_n = [nx]/ n$ for $xϵR$ where $[·]$ denotes the integer part. Then ${x_n}$
(a) converges to $x$.
(b) converges but not to $x$.
(c) does not converge
(d) oscillates.

I think (a) is correct as $\lim_{n \to \infty}[n]/n=1$. Am I right?

share|improve this question
    
answer should depend on x. –  i. m. soloveichik Nov 23 '12 at 14:20

3 Answers 3

up vote 1 down vote accepted

You are right since for $x=1$ you get $\displaystyle{\lim_{n \to \infty}\frac{[xn]}{n}=\lim_{n \to \infty}\frac{[n]}{n}=1=x}$.
Therefore you can eliminate (b),(c) and (d) and the correct answer is (a).
For the proof use that $$nx-1<[nx]\leq nx\Rightarrow x-\frac{1}{n}<\frac{[nx]}{n}\leq x, \ \forall n \in \mathbb{N}.$$

share|improve this answer

We have $|\,\lfloor nx\rfloor -nx\,|<1$, which implies $$ \left|\frac{\lfloor nx\rfloor}n - x\right|<\frac1n. $$ So $x_n$ always converges to $x$, for any $x\in\mathbb R$.

share|improve this answer

The answer is (a), but your reasoning is not correct: you cannot factor out $x$ like that. What you can do is observe that

$$0\le x-\frac{\lfloor nx\rfloor}n=\frac{nx-\lfloor nx\rfloor}n<\frac1n$$

and take limits as $n\to\infty$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.