Is this proof correct: $\liminf {\frac{\sin(\frac{\pi n}{8})n^n}{n!}} = 0$?

Question

Evaluate a limit of a sequence: $\liminf {\frac{\sin(\frac{\pi n}{8})n^n}{n!}}$.

Proof: $\liminf {\frac{\sin(\frac{\pi n}{8})n^n}{n!}} = 0$. Since a product of convergent sequence and a bounded sequence is convergent, $\liminf {\frac{sin(\frac{\pi n}{8})n^n}{n!}} = \lim{\frac{n^n}{n!} \cdot \liminf{sin(\frac{\pi n}{8})}}$.

So $\liminf {\frac{\sin(\frac{\pi n}{8})n^n}{n!}} = 0 \cdot (-1) = 0$.

Is this proof correct with all steps properly justified?

Answer 1

No, your proof is not correct.

Consider the subsequence $n_k:=12+16k$ for $k\geq 0$, then $$\sin(\pi n_k/8)=\sin\left(\frac{3\pi}{2}+2k\pi\right)=-1.$$ Moreover for $n\geq 2$ $$\frac{n^n}{n!}=n\cdot\prod_{j=2}^n\frac{n}{j}\geq n.$$ (which means that the sequence $\frac{n^n}{n!}$ is unbounded!). Hence $$\liminf_{n\to+\infty} {\frac{\sin(\frac{\pi n}{8})n^n}{n!}} \leq -\lim_{k\to+\infty} \frac{{n_k}^{n_k}}{n_k!}\leq -\lim_{k\to+\infty} n_k=-\infty.$$