Answer
Converges
Work Step by Step
Let us consider $a_n=\dfrac{2+(-1)^n}{1.25^{n}}$
Apply direct comparison test.
This implies that $\Sigma_{n=1}^\infty \dfrac{2+(-1)^n}{(1.25)^{n}} \leq \Sigma_{n=1}^\infty \dfrac{3}{(1.25)^{n}}=(3) \Sigma_{n=1}^\infty (\dfrac{1}{1.25})^{n} $
Here, the series $ \Sigma_{n=1}^\infty (\dfrac{1}{1.25})^{n} $ shows a geometric convergent series.
Thus, the series converges by the the direct comparison test.