checking a sequence cauchy or not or convergent Could any one help me to show whether the following are cauchy or convergent?
$(1)$ $\{1/n\}$ in $(0,1)$
$(2)$ $\{1/n^p\}$ in $\mathbb{R}, p>0$
$(3)$ $\{n^{\frac{1}{n}}\}$ in $\mathbb{R}$
I loosely understand that $\frac{1}{n}\to 0\notin(0,1)$ and $\exists m,n \in\mathbb{N}\ni |x_n-x_m|=|\frac{1}{n}-\frac{1}{m}|<\epsilon\forall m,n>N$
thanks for helping 
 A: You can use the following proposition:

If a sequence is convergent, then it is also Cauchy.

$(1)$ Since $\lim_{n\to\infty}\frac{1}{n} = 0$, the sequence $\left(\frac{1}{n}\right)_{n=1}^\infty$ is convergent in $\mathbb{R}$, so it is Cauchy. However, $0 \notin (0, 1)$, so $\left(\frac{1}{n}\right)_{n=1}^\infty$ is not convergent in $(0, 1)$, but it still is Cauchy.
$(2)$ $\left(\frac{1}{n^p}\right)_{n=1}^\infty$ is convergent in $\mathbb{R}$ and $\lim_{n\to\infty}\frac{1}{n^p} = 0$, as we can show:
Let $\varepsilon > 0$ and let $\lceil\cdot\rceil$ denote the ceiling function. For $n \in\mathbb{N}, n \geq 1+\left\lceil\frac{1}{\sqrt[p]{\varepsilon}}\right\rceil$ we have:
$$\left|\frac{1}{n_p} - 0\right| = \frac{1}{n^p} \leq \frac{1}{\left(1+\left\lceil\frac{1}{\sqrt[p]{\varepsilon}}\right\rceil\right)^p} < \frac{1}{\left(\frac{1}{\sqrt[p]{\varepsilon}}\right)^p} = \varepsilon$$
This implies that $\left(\frac{1}{n^p}\right)_{n=1}^\infty$ is also Cauchy.
$(3)$ $\left(\sqrt[n]{n}\right)_{n=1}^\infty$ is convergent in $\mathbb{R}$ and $\lim_{n\to\infty} \sqrt[n]{n} = 1$. To demonstrate this, let $\varepsilon > 0$. Let $n \geq 2 + \left\lceil\frac{2}{\varepsilon^2}\right\rceil$. We have, using the binomial theorem:
$$n = \left(\sqrt[n]{n}\right)^n = \left(1 + \left(\sqrt[n]{n}-1\right)\right)^n = \sum_{k=0}^n\binom{n}{k}\left(\sqrt[n]{n}-1\right)^k \geq \binom{n}{2}\left(\sqrt[n]{n}-1\right)^2 = \frac{n(n-1)}{2}\left(\sqrt[n]{n}-1\right)^2$$
This implies:
$$|\sqrt[n]{n} - 1| \leq \sqrt{\frac{2}{n-1}} \leq \sqrt{\frac{2}{\left(2 + \left\lceil\frac{2}{\varepsilon^2}\right\rceil\right)-1}} < \sqrt{\frac{2}{\frac{2}{\varepsilon^2}}} = \varepsilon$$
Since it is convergent, $\left(\sqrt[n]{n}\right)_{n=1}^\infty$ is also Cauchy.
