Answer
$\left\{ {{b_n}} \right\}$ diverges.
Work Step by Step
We write ${b_n} = \frac{{n!}}{{{\pi ^n}}}$ as a product of $n$ factors:
$\frac{{n!}}{{{\pi ^n}}} = \frac{1}{\pi }\cdot\frac{2}{\pi }\cdot...\cdot\frac{n}{\pi }$.
By (iv) of Theorem 2 we get
$\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{{\pi ^n}}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{\pi }\cdot\frac{2}{\pi }\cdot...\cdot\frac{n}{\pi }} \right) = \frac{1}{\pi }\cdot\frac{2}{\pi }\cdot...\cdot\frac{1}{\pi }\left( {\mathop {\lim }\limits_{n \to \infty } n} \right)$.
Since the sequence $\left\{ n \right\}$ diverges, so $\left\{ {{b_n}} \right\}$ diverges.
