Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{2+(-1)^n}{1.25^{n}}$
Here, $\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} $
we see that $3 \Sigma_{n=1}^\infty (\dfrac{1}{1.25})^{n} $ is a geometric convergent series.
Hence, the given series converges by the direct comparison test.