The set: $$\Big\{\frac{1}{n} + (-1)^{n} \mid n \in \mathbb{N}\Big\}$$ So I know that if I plotted this graph it would be full of discontinuities as n changes from odd to even but I'm not sure how I can find the supremum and infimum without using any limits (or calculus but I don't think that's important here anyways, correct me if I'm wrong). I think that the supremum is going to occur when $n$ is even and n is as small as possible. So in this case I think the supremum would be $\frac{3}{2}$, when $n$ is equal to $2$. I think the infimum would be $-1$ as when $n$ tends to infinity and is odd it will output $-1$.
I'd really appreciate if someone could tell me how to do this sort of question more rigorously or if I've made a mistake please point it out. I would greatly appreciate it. Thank you
Consider the case where $n$ is odd. Then, $\frac{1}{n}+(-1)^{n} = \frac{1}{n}-1$. Thus, the infimum for the sequence of odd $n$ is $-1$ as $n\rightarrow \infty$, and the supremum is $0$ at $n=1$.
Now, consider the case where $n$ is even. Then, $\frac{1}{n}+(-1)^{n} = \frac{1}{n} + 1$. Thus, the infimum for the sequence of even $n$ is $1$ as $n\rightarrow\infty$, and the supremum is $\frac{3}{2}$ at $n=2$.
Thus, the infimum for the entire set is $-1$, and the supremum is $\frac{3}{2}$. $\blacksquare$