Determine the convergence or divergence of f(n) = (n2/3 sin (n!)) / (n+1)

Consider the function $f(n)$ defined by

$f(n) = \frac{n^{\frac{2}{3}} \sin (n!)}{n+1}.$

Determine whether the sequence $\{ f(n) \}$ converges or diverges, and if it converges find the limit.

We divide the numerator and denominator of $f(n)$ by $\frac{1}{n}$ ,

$\begin{align*} f(n) &= \frac{n^{\frac{2}{3}} \sin (n!)}{n+1} \\[9pt] &= \frac{ \frac{1}{n^{\frac{1}{3}}} \sin (n!) }{1 + \frac{1}{n}}. \end{align*}$

Since $|\sin (n!)| \leq 1$ for all $n$ we have

$\lim_{n \to \infty} \frac{1}{n^{\frac{1}{3}}} \sin (n!) = 0.$

Since

$\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) = 1$

we then have that both limits in the quotient exist, and the limit of the denominator is nonzero; hence,

$\lim_{n \to \infty} f(n) = \frac{0}{1} = 0.$

Therefore, the sequence $\{ f(n) \}$ converges to the limit 0.

Stumbling Robot