Answer
The sequence is divergent.
Work Step by Step
$\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty} \frac{1 \times 3 \times 5...(2n-1)}{n!}$
$=\lim\limits_{n \to \infty}\frac{1 \times 3 \times 5...(2n-1)}{1 \times 2\times 3...n}$
$=\lim\limits_{n \to \infty}\frac{3}{2} \times \frac{5}{3} \times \frac{7}{4}... (2-\frac{1}{n})$
We know that $\lim\limits_{n \to \infty}\frac{1}{n}=0$ and we know that since the numerator increases at a faster rate than the denominator, we can tell that the product $a_{n}$ will increase without bound to $\infty$.
Thus, the sequence is divergent.
