I want to show that $\displaystyle S_n = \begin{cases} 3-\sqrt{n}, & \mbox{if n is even} \\ -\frac{\sqrt{n}}{3}, & \mbox{if n is odd } \end{cases}$ converges to $-\infty$ as $n \to \infty$.
I've tried as follows:
Let $m>0$. Take $N =\lceil (3-m)^2\rceil+1$. Then $n > N$ gives:
$n > (3-m)^2 \implies \sqrt{n} >3-m \implies -\sqrt{n} < m-3 \implies 3-\sqrt{n} <m$ if $n$ is even
$\displaystyle n > (3-m)^2 \implies \sqrt{n} >3-m < -\frac{\sqrt{n}}{3} < m/3-1 \implies 3-\sqrt{n} <m$ if $n$ is odd.
Does this work?
There are several things that confuse me here. For one $-\sqrt{n}/3$ is always less than $0$, so would taking any $m>0$ suffice? Second, do I need to use absolute value sign somewhere?