Calculus (3rd Edition)

No, Ralph is not right because the series $\Sigma_{n=1}^\infty \frac{1}{n}$ diverges and the series $\Sigma_{n=1}^\infty \frac{e^{-n}}{n}=\Sigma_{n=1}^\infty \frac{1}{ne^n}$ is smaller than $\Sigma_{n=1}^\infty \frac{1}{n}$. The fact that the larger series diverges does not mean that the smaller one diverges.