Calculus: Early Transcendentals (2nd Edition)

$\Sigma_{k=1}^\infty2k^\frac{-3}{2}=2\Sigma_{k=1}^\infty\frac{1}{k^\frac{3}{2}}$ $p=\frac{3}{2}\gt1$, therefore $\Sigma_{k=1}^\infty\frac{1}{k^\frac{3}{2}}$ converges as does $2\Sigma_{k=1}^\infty\frac{1}{k^\frac{3}{2}}$