# Limit of a sequence $\frac{n^{2/3} \sin(n!)}{n+1}$

I am having a problem calculating the following limit:

$$\lim\limits_{n \to \infty} \frac{n^{2/3} \sin(n!)}{n+1}$$

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Hint: $|\sin x|\le 1$. –  André Nicolas Nov 14 '12 at 15:17

Think about what's happening as $n\to \infty$: $n^{2/3}$ and $n$ are both growing without bound, but $\sin n!$ is bounded above by $1$, since the sine function can only take on values between $-1$ and $1$. So as $n$ increases, the $\sin n!$ term becomes inconsequentially small compared to the others. Then

$$\displaystyle \lim_{n\to\infty} \frac{n^{2/3}\sin(n!)}{n+1} = \lim_{n\to\infty} \frac{n^{2/3}}{n+1}$$

Now since $n^{2/3}$ is a smaller power than $n$, it grows more slowly towards infinity. Thus

$$\lim_{n\to\infty} \frac{n^{2/3}}{n+1}= 0.$$

So the limit of the sequence is $0$.

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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 $a_nb_n\xrightarrow [n\to\infty]{} 0\,$

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