The series is alternating, suggesting the Leibniz test. For this, $\left\{\ln \left(n\right)\/\sqrt{n}\right\}$ must eventually become a monotone decreasing sequence with limit zero. Consequently, the following calculations are executed.

$\underset{n\to \infty }{lim}\ln \left(n\right)\&sol;\sqrt{n}$ = $0$  and   $\frac{d}{\mathrm{dx}}\left(\ln \left(x\right)\&sol;\sqrt{x}\right)\&equals;\frac{2-\ln \left(x\right)}{2 x^{3\&sol;2}}<0$ for $\ln \left(x\right)\&gt;2$, or $x\&gt;e^2\doteq 7.4$ 

The conditions of the Leibniz test apply, and by that test, the series converges conditionally.

To test for absolute convergence, apply the Integral test: ${\int }_8^{\infty }\frac{\ln \left(x\right)}{\sqrt{x}} \mathit{\&DifferentialD;}x$ = $\mathrm{\&infin;}$.

By this test, the series of absolute values does not converge, so the series is not absolutely convergent.

A modicum of algebra puts the derivative into the form $\frac{2-\ln \left(x\right)}{2 x^{3\&sol;2}}$, from which it can be seen that the derivative is negative for $2-\ln \left(x\right)<0$, or $x\&gt;e^2\doteq 7.4$. (This is consistent with Figure 8.3.6(a).) Since the criteria of the test are satisfied, it follows that the series converges conditionally.