Is $\frac{1}{\sqrt x}$ a converging or diverging sequence?

Question

Suppose that $\{\epsilon\}$ is a sequence of positive reals converging to 0. Check whether the set $\{n\epsilon \}$ is bounded. I am looking for the answer of this question.

Answer 1

The set $\{n\varepsilon_n\}$ need not be bounded. Consider $\varepsilon_n=\frac1{\log n}.$ $$\lim_{n\to\infty}\frac n{\log n}=\infty.$$

Answer 2

Well the would mean $\epsilon_n$ must be of big O of $1/n$.theres thousands of sequences converging to zero but not of order $1/n$

Answer 3

Title: yes, $\lim\limits_{x \rightarrow \infty} \frac{1}{\sqrt x}$ converges, as it is bounded below and decreasing. Second question: No, $\{ \epsilon \}$ convergent does not imply $\{ n\epsilon \}$ converges. Consider for example, $\{\frac{1}{\sqrt n}\}$, which converges, but $\{n \frac{1}{\sqrt n}\} = \{\sqrt n\}$ does not.

Answer 4

More generally, take $\epsilon_n=\frac{1}{n^ \alpha}$ with $\alpha<1$.