## Calculus 8th Edition

$\Sigma_{n=1}^{\infty} \frac{\sqrt n+4}{n^{2}}=\Sigma_{n=1}^{\infty} \frac{\sqrt n}{n^{2}}+\Sigma_{n=1}^{\infty} \frac{4}{n^{2}}$ Here, $\Sigma_{n=1}^{\infty} \frac{\sqrt n}{n^{2}} =\Sigma_{n=1}^{\infty} \frac{1}{n^{3/2}}=$convergent by p-series and $\Sigma_{n=1}^{\infty} \frac{4}{n^{2}}$= convergent by p-series. Hence, the given series is convergent.