According to the Leibniz' test, the alternating series $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7} -\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+-\ldots$$ is convergent and has a positive sum ($= \ln{2}$ ; see the natural logarithm). Denote it by $S$ . We can group pairwise its terms and multiply each term by $\frac{1}{2}$ getting the two series $S = (1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{7} -\frac{1}{8})+(\frac{1}{9}-\frac{1}{10})+\ldots,$

$\frac{1}{2}S = \frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-+\ldots.$

Then we add these two series termwise getting the sum

$1\frac{1}{2}S = 1+\frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\frac{1}{7} -\frac{2}{8}+\frac{1}{9}+\frac{1}{11}-\frac{2}{12}+\ldots.$

Hence, this last series contains exactly the same terms as the original, but its sum is fifty percent greater. This is possible because the original series is not absolutely convergent: the series which is formed of the absolute values of its terms is the divergent harmonic series.

P. S. - For justification of the used manipulations of the series, see the parent entry.