I wish to show if $S_n = \begin{cases} 2e^{-n}, & \mbox{if n is even } \\ -\frac{3}{n}, & \mbox{if n is odd } \end{cases}$ then we have $S_n \to 0$ as $n \to \infty$.
Does it suffice to consider the the odd/even terms of the sequence separately? e.g.
If $n$ is even, take $ N > \log(2/\epsilon)$. Then $n > N \implies n > \log(2/3) \implies |2e^{-n}|< \epsilon$
And if $n$ is odd, then take $N > -\frac{3}{\epsilon}$. Then $n> N \implies n > -\frac{3}{\epsilon} \implies |-\frac{3}{n}| < \epsilon$.
That's right. Also notice that for $n$ being odd, we must have $$N>{3\over \epsilon}$$not $N>-{3\over \epsilon}$. The general $N$ for both even and odd terms then would become$$N>\max\left\{{3\over \epsilon} , \log{2\over \epsilon} \right\}$$