Answer
Converges absolutely
Work Step by Step
Let ${a_k} = {\left( { - \dfrac{3}{5}} \right)^k} = {\left( { - 1} \right)^k}{\left( {\dfrac{3}{5}} \right)^k}$.
Evaluate the limit:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{\left| {{a_{k + 1}}} \right|}}{{\left| {{a_k}} \right|}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{{\left( {\dfrac{3}{5}} \right)}^{k + 1}}}}{{{{\left( {\dfrac{3}{5}} \right)}^k}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{3}{5} = \dfrac{3}{5}$
The series converges absolutely by the ratio test for absolute convergence because $\rho = \dfrac{3}{5} \lt 1$. Hence, it converges.
