Answer
$\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{3^n} - {2^n}}}$ converges.
Work Step by Step
Write ${a_n} = \dfrac{1}{{{3^n} - {2^n}}}$ and ${b_n} = \dfrac{1}{{{3^n}}}$.
We apply the Limit Comparison Test:
$L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{3^n}}}{{{3^n} - {2^n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{1 - \dfrac{{{2^n}}}{{{3^n}}}}} = 1$
Since $L \gt 0$ and $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{3^n}}}$ converges (because it is a geometric series with ratio $r = \dfrac{1}{3} \lt 1$), by the Limit Comparison Test, $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{3^n} - {2^n}}}$ converges.
