Answer by Akash Gaur for Consider the sequence ${x_n}$ defined by $x_n =...
$$x_n=\frac{\lfloor nx \rfloor}{n}=\frac{nx-\{nx\}}{n}=x-\frac{\{nx\}}{n}$$$$0\leq\{nx\}<1\implies\lim_{n\to\infty}x_n=x$$
View ArticleAnswer by Brian M. Scott for Consider the sequence ${x_n}$ defined by $x_n =...
The answer is (a), but your reasoning is not correct: you cannot factor out $x$ like that. What you can do is observe that $$0\le x-\frac{\lfloor nx\rfloor}n=\frac{nx-\lfloor...
View ArticleAnswer by Martin Argerami for Consider the sequence ${x_n}$ defined by $x_n =...
We have $|\,\lfloor nx\rfloor -nx\,|<1$, which implies$$\left|\frac{\lfloor nx\rfloor}n - x\right|<\frac1n.$$So $x_n$ always converges to $x$, for any $x\in\mathbb R$.
View ArticleAnswer by P.. for Consider the sequence ${x_n}$ defined by $x_n = [nx]/ n$
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...
View ArticleConsider the sequence ${x_n}$ defined by $x_n = [nx]/ n$
Consider the sequence ${x_n}$ defined by $x_n = [nx]/n$ for $x\in\mathbb R$ where $[·]$ denotes the integer part. Then ${x_n}$(a) converges to $x$.(b) converges but not to $x$.(c) does not converge(d)...
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