I was looking at a practice exam for an analysis test and came across this question.
$$x_n := e^{\large n\sin(n\pi/4)}$$
i. Is this sequence bounded? ii. Does there exist a convergent subsequence? iii. Do lim sup and lim inf exist as finite numbers
For (i.), intuition says it is unbounded for there can always be a greater positive power to which to raise e.
For (ii.) intuition says there does exist a convergent subsequence for when $\sin((n\pi)/4) = 0, e ^ {n\sin((n\pi)/4)}$ will be a sequence of 1s, hence it converges to 1.
For (iii.) I am not quite sure how to articulate lim sup and lim inf.
I'm not looking for any explicit answer, but more guidance as to how one approaches such a problem.
Sorry, I'm new to the syntax here and will try to make the sequence definition more readable
i. If $n$ is of the form $8k +2$ with $k\in \mathbb{N}$ then $$x_n = e^{n\sin(2k\pi + \frac \pi 2)} = e^n,$$ therefore for each positive constant $C$ it is possible to find a rank $n$ such that $|x_n| > C$ (because $\lim_{n\rightarrow + \infty} e^n = +\infty$).
ii. Consider the subsequence $(x_{k(n)})$ where $k(n) = 4n$, we have $$x_{k(n)} = e^{4n\sin(\pi)} = e^{4n\cdot 0} = 1,$$ therefore the subsequence $(x_{k(n)})$ converges to $1$.
iii. From i. it is clear that $\lim\sup x_n = +\infty$. And from ii. it is known that $$\inf_{n\ge k} x_n \le 1$$ for all $k \in \mathbb{N}$ and because an exponential is always positive we have $$0 \le \inf_{n\ge k} x_n \le 1$$ for all $k \in \mathbb{N}$ therefore $\lim\inf x_n$ is a finite number between $0$ and $1$.