The series is alternating, suggesting the Leibniz test. For this, $\left\{1\/ \ln \left(n\right)\right\}$ must eventually become a monotone decreasing sequence with limit zero. Consequently, the following calculations are executed.

$\underset{n\to \infty }{lim}1\&sol; \ln \left(n\right)$ = $0$  and   $\frac{d}{\mathrm{dx}}\left(1\&sol; \ln \left(x\right)\right)\&equals;-\frac{1}{x \ln {\left(x\right)}^2}<0$ for $x\&gt;1$

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

To test for absolute convergence, apply the Integral test: ${\int }_2^{\infty }\frac{1}{\ln \left(x\right)} \mathit{\&DifferentialD;}x$ = $\mathrm{\&infin;}$.

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

The following two calculations support the use of the Integral test.

The derivative is negative for $x\&gt;1$. (This is consistent with Figure 8.3.14(a).) Since the criteria of the test are satisfied, it follows that the series does not converge absolutely. The Leibniz test applies, and by it, this alternating series is seen to converge conditionally.