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

Consider $a_n=b_n=(\dfrac{1}{2})^n$ Let $A=\Sigma_{n=1}^{\infty}a_n=\Sigma_{n=1}^{\infty}(\dfrac{1}{2})^n=1$ and $B=\Sigma_{n=1}^{\infty}b_n=\Sigma_{n=1}^{\infty}(\dfrac{1}{2})^n=1$ and $\Sigma_{n=1}^{\infty}(a_nb_n)=\Sigma_{n=1}^{\infty}(\dfrac{1}{4})^n=\dfrac{1}{3} \ne AB$