Cauchy sequence question. Show that $\frac{(-1)^n}{\sqrt{n}}$ is Cauchy.
All up until the * is my work, the rest is from the solution, of which i don't understand.
Let $m > n$, then
$|\frac{(-1)^m}{\sqrt{m}} - \frac{(-1)^n}{\sqrt{n}}| \le |\frac{(-1)^m}{\sqrt{m}}| +|\frac{(-1)^n}{\sqrt{n}} = |\frac{1}{\sqrt{m}} + \frac{1}{\sqrt{n}}| < \frac{2}{\sqrt{n}}$.
*
so if for a given $\epsilon$ > 0, we choose $N_1 \ge \frac{2}{\epsilon} $ then...
Is this because  $\frac{2}{\sqrt{n}}$ is a decreasing sequence for n? I just cant quite make the connection here, really cloudy.
 A: Solution 1. Since $\lim_{n\rightarrow\infty}\frac{(-1)^n}{\sqrt{n}}=0$ and every convergent sequence is a Cauchy sequence, $\left\{\frac{(-1)^n}{\sqrt{n}}\right\}$ is a Cauchy sequence.
Solution 2. We have
$$
\left|\frac{(-1)^n}{\sqrt{n}}-\frac{(-1)^m}{\sqrt{m}}\right|\leq\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{m}}\rightarrow 0 \quad\text{as}\quad m,n\rightarrow\infty
$$
It follows that
$$
\lim_{m,n\rightarrow\infty}\left|\frac{(-1)^n}{\sqrt{n}}-\frac{(-1)^m}{\sqrt{m}}\right|=0.
$$
Solution 3. Let $\varepsilon>0$. Choosing $N_0\in\mathbb{N}$ such that 
$$
N_0>\frac{4}{(\varepsilon)^2}.
$$
Then for $m,n\in\mathbb{N}$ such that $m\geq N_0, n\geq N_0$ we have
$$
\frac{1}{\sqrt{m}}\leq\frac{1}{\sqrt{N_0}}<\frac{\varepsilon}{2}, \quad \frac{1}{\sqrt{n}}\leq\frac{1}{\sqrt{N_0}}<\frac{\varepsilon}{2}.
$$
It follows that
$$
\left|\frac{(-1)^n}{\sqrt{n}}-\frac{(-1)^m}{\sqrt{m}}\right|\leq\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{m}}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.
$$
A: To answer your question about the choice of $N_1$...
You have already arrived at 
$$ |\frac{(-1)^m}{\sqrt{m}} - \frac{(-1)^n}{\sqrt{n}}| \le \frac{2}{\sqrt{n}}  $$
So given an $\epsilon \gt 0$ since there are arbitrarily large natural numbers, there is $N_1 \in \Bbb N$ such that $ N_1  \gt \dfrac{ 4}{\epsilon^2} \iff  \epsilon \gt  \dfrac{2}{\sqrt{N_1}}$. 
Now let $m, n$ be arbitrary natural numbers greater than $N_1$. Without loss of generality you may suppose that $m \ge n$ as well. Then, 
$$ |\frac{(-1)^m}{\sqrt{m}} - \frac{(-1)^n}{\sqrt{n}}| \le \frac{2}{\sqrt{n}} \lt \dfrac{2}{\sqrt{N_1}} \lt \epsilon  $$
And that is why this is a Cauchy sequence. 
