I'm trying to show that $(a_{n}) \to 0$ then $(\frac{1}{(a_{n})}) \to \infty$ is false.
As a counter example I have $(a_{n}) = \frac{(-1)^{n}}{n}$. This converges to 0, but for the reciprocal we have that $\frac{1}{a_{2n}} \to \infty$ and $\frac{1}{a_{2n-1}} \to -\infty$
I'm struggling to formally show that $\frac{1}{a_{2n}} \to \infty$.
Attempt:
We want to show for all $C > 0$ there is some $N$ such that $\frac{1}{a_{2n}} > C$ for all $n \geq N$.
So let $C >0$ be arbitrary. If we can solve for $n$ in the following, I can choose $N$ correctly. So I need to solve for $n$ in $\frac{2n}{(-1)^{2n}} > C$
But the $(-1)^{2n}$ is confusing me because when I take logs and separate I end up with $\ln(2n) - 2n\ln(-1) > \ln(C)$, but since this is real analysis we can't have $\ln(-1)$...
You are complicating things too much.
Simply noting what happens for $n$ odd, and $n$ even suffices.
Note that : $$ a_n = \begin{cases} \frac 1n & n \text{ even} \\ \frac {-1}n & n \text{ odd} \end{cases} $$
and therefore : $$ \frac1{a_n} = \begin{cases} n & n \text{ even} \\ -n & n \text{ odd} \end{cases} $$
Now it is almost obvious that $\frac 1{a_{2k}} \to \infty$ and $\frac 1{a_{2k+1}} \to - \infty$ : the formal proof is an exercise in understanding the definition, really.