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

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Consider $a_n=b_n=(\dfrac{1}{2})^n$ The given series $\Sigma_{n=1}^{\infty}a_n=\Sigma_{n=1}^{\infty}b_n=\Sigma_{n=1}^{\infty} (\dfrac{1}{2})^n=1$ and $\Sigma_{n=1}^{\infty}(\dfrac{a_n}{b_n})=\Sigma_{n=1}^{\infty}(1)$, which diverges.