Limit of a sequence $\frac{n^{2/3} \sin(n!)}{n+1}$ I am having a problem calculating the following limit:
$$\lim\limits_{n \to \infty} \frac{n^{2/3} \sin(n!)}{n+1}$$
Please help.
Thank you in advance

Since
$$0\leq\frac{n^{2/3}}{n+1}\leq\frac{n^{2/3}}{n}=\frac{1}{\sqrt[3]n}\xrightarrow[n\to\infty]{} 0$$
and also $\,|\sin (n!)|\leq 1\,$ ,as you can use the following easy-to-prove lemma:
Lemma: If $\,a_n\xrightarrow [n\to\infty]{} 0\,$ and $\,\{b_n\}\,$ bounded, then $a_nb_n\xrightarrow [n\to\infty]{} 0\,$
