## University Calculus: Early Transcendentals (3rd Edition)

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.