So, I don't know how correctly show that,
$c(n) := \left(\frac{(-1)^n}{\sqrt{n}}\right)\to 0\qquad\text{ as}\quad n\to\infty.$
Should I do this with limes or by $\forall \epsilon > 0, \exists N(\epsilon)$ such that $\forall n >= N(\epsilon) :| c(n) - c | < \epsilon$?
Another way to prove it (without knowing your background, you may or may not know all these tricks) is to consider the alternating series:
$$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n}$$
Using the Alternating Series Test, we have $\frac{1}{\sqrt n} \rightarrow 0$ and $\frac{1}{\sqrt n} > \frac{1}{\sqrt{n+1}}$. Thus the series converges, so for the series to converge, we must have $\frac{(-1)^n}{\sqrt n} \rightarrow 0$
Standard argument is: For any given $\epsilon > 0$, choose $N = \lfloor{\dfrac{1}{\epsilon^2}\rfloor}+1\Rightarrow \forall n \geq N, |c_n| = \left|\dfrac{1}{\sqrt{n}}\right| < \epsilon$, and the answer follows.
