Answer
$\left\{ {{a_n}} \right\}$ diverges.
Work Step by Step
We have
$\mathop {\lim }\limits_{n \to \infty } \frac{{3 - {4^n}}}{{2 + 7\cdot{3^n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{3 - {4^n}}}{{2 + 7\cdot{3^n}}}\frac{{1/{3^n}}}{{1/{3^n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{3}{{{3^n}}} - {{\left( {\frac{4}{3}} \right)}^n}}}{{\frac{2}{{{3^n}}} + 7}}$
$ = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{3}{{{3^n}}}}}{{\frac{2}{{{3^n}}} + 7}} - \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\frac{4}{3}} \right)}^n}}}{{\frac{2}{{{3^n}}} + 7}} = \frac{{\mathop {\lim }\limits_{n \to \infty } \frac{3}{{{3^n}}}}}{{\mathop {\lim }\limits_{n \to \infty } \frac{2}{{{3^n}}} + 7}} - \frac{{\mathop {\lim }\limits_{n \to \infty } {{\left( {\frac{4}{3}} \right)}^n}}}{{\mathop {\lim }\limits_{n \to \infty } \frac{2}{{{3^n}}} + 7}} = \frac{0}{7} - \frac{\infty }{7} = - \infty $
So, $\left\{ {{a_n}} \right\}$ diverges.
