Answer
$\left\{ {{b_n}} \right\}$ diverges.
Work Step by Step
We have
$\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n}{n^3} + {2^{ - n}}}}{{3{n^3} + {4^{ - n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n}{n^3} + {2^{ - n}}}}{{3{n^3} + {4^{ - n}}}}\frac{{1/{n^3}}}{{1/{n^3}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n} + {2^{ - n}}/{n^3}}}{{3 + {4^{ - n}}/{n^3}}}$
$\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n} + \frac{1}{{{2^n}{n^3}}}}}{{3 + \frac{1}{{{4^n}{n^3}}}}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( { - 1} \right)}^n}}}{{3 + \frac{1}{{{4^n}{n^3}}}}} + \frac{{\frac{1}{{{2^n}{n^3}}}}}{{3 + \frac{1}{{{4^n}{n^3}}}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( { - 1} \right)}^n}}}{{3 + \frac{1}{{{4^n}{n^3}}}}}} \right) + 0$
$\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( { - 1} \right)}^n}}}{{3 + \frac{1}{{{4^n}{n^3}}}}}} \right)$
Since the term on the right-hand side bounce back and forth with positive and negative values but never settle down to approach a limit, so $\left\{ {{b_n}} \right\}$ diverges.
