Find $\liminf\limits_{n\to\infty} (x_{n})$ and $\limsup\limits_{n\to\infty}(x_{n})$ for a sequence $x_{n}=1-n\sin\frac{n\pi}{4}$

Question

Subsequence $a_{n_{1}}=\sin\frac{n\pi}{4}$ is bounded ($[-1,1]$), and a subsequence $a_{n_{2}}=n$ is bounded below. We can find cluster points for the first subsequence $C_{1}=\{-1,1\}$. For the second subsequence, $$\lim\limits_{n\to\infty}n=\infty$$

Does this means that the second subsequence doesn't have cluster points?

If $$-1\le \sin\frac{n\pi}{4}\le 1$$ than $$-n \le n\sin\frac{n\pi}{4}\le n$$

so the cluster points for subsequence $a_{n_{3}}=n\sin\frac{n\pi}{4}$ are $$C_{2}=\{-n,n\}$$

My solutions are $$\liminf\limits_{n\to\infty}\,(x_{n})=-n$$ $$\limsup\limits_{n\to\infty}\,(x_{n})=n$$

Is this right?

Answer 1

The sequence $a_n$ has more cluster points than just $\{-1,1\}$, namely $\{-1,-\sqrt 2/2,0\sqrt 2/2,1\}$ (in fact, it is periodic). From the values where $-1$ or $1$ is achieved, we see that $1-n a_n$ is not bounded from below, nor from above, so $\liminf x_n = -\infty$, $\limsup x_n=+\infty$. On the other hand, the subsequence where $a_n=0$ leads to a cluster point of $x_n$, namely $1$.