I want to show that $x_n = \frac{(-1)^n}{\sqrt{n}}$ is a Cauchy sequence. It is not given that this sequence is convergent.
To that I end, I need to find an $N$ in the natural numbers such that $n,m > N \Rightarrow |x_n - x_m|<\epsilon$. This would work if $x_n,x_m<\frac{\epsilon}{2}$.
The problem is that, after doing the algebra, the $N$ I came up with is not necessarily a natural number; I got $N=(\frac{2}{\epsilon})^2$.
Any advice on what I might be missing would be greatly appreciated.
First, it's good to be careful: you want $\lvert x_n \rvert < \frac{\varepsilon}{2}$ for $n > N$, not just $x_n < \frac{\varepsilon}{2}$.
You presumably proved that $\lvert x_n \rvert < \frac{\varepsilon}{2}$ for $n > (2/\varepsilon)^2$. We want to replace $(2/\varepsilon)^2$ in this statement by a natural number so that the claim still holds. There are many ways to do this, but the easiest is probably to choose $N = \lceil (2/\varepsilon)^2 \rceil$.