## Calculus: Early Transcendentals 8th Edition

As a summation, the series can be written as $\Sigma_{n=1}^{\infty}\frac{1}{n\sqrt n}=\Sigma_{n=1}^{\infty}\frac{1}{n^{3/2}}$, which is a convergent p-series with $p=\frac{3}{2}\gt 1$. Hence, the given series is convergent.