Since $\sqrt{n^3\+5}$ behaves as $n^{3\/2}$ for large $n$, comparison to the convergent $p$-series $\mathrm{\Σ} n^{-3\/2}$ suggests the calculation

$\underset{n\to \infty }{lim}\frac{\left(\frac{1}{\sqrt{n^3\+5}}\right)}{n^{-3\/2}}$ = $\underset{n\to \infty }{lim}\frac{n^{3\/2}}{\sqrt{n^3\+5}}\=1$  

from which, by part (1) of the Limit-Comparison test, the convergence of the given series is established. Since it is a convergent series with positive terms, the series converges absolutely.

As per the Mathematical Solution above, the absolute convergence of the given series can be established by the Limit-Comparison test, by a comparison to the convergent $p$-series $\sum _{n\=1}^{\infty }\frac{1}{n^{3\/2}}$. Absolute convergence can also be established by the Integral test, but the integration is not elementary.

Since the integral converges, the series converges absolutely (as it is a series of positivel terms).