Rearranging a conditionally convergent series

Idea. Add the series $S$ to the series $\frac S2$, in which the term $\frac{(-1)^{k+1}}{2k}$ is placed at position $2k$ and zeros elsewhere:
$$\frac S2=0+\frac12+0-\frac14+0+\frac16+0-\frac18+\cdots$$
Two convergent series can be added term by term.

Addition.
- position $1$: $1+0=1$
- position $2$: $-\frac12+\frac12=0$
- position $3$: $\frac13$
- position $4$: $-\frac14-\frac14=-\frac12$
- position $5$: $\frac15$
- position $6$: $-\frac16+\frac16=0$
- position $7$: $\frac17$
- position $8$: $-\frac18-\frac18=-\frac14$

In general the even positions alternate between $0$ and $-\frac1{2k}$ for $k$ even, and the odd positions are $\frac1{2k+1}$.

Removing zeros. Deleting zero terms from a convergent series does not change its sum. We obtain
$$\frac32S=1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16+\cdots$$
which is exactly the rearrangement (every positive term $\frac1{2k+1}$ and every negative term $-\frac1{2k}$ appears once).

Conclusion. The rearranged series sums to $\frac32\ln2\approx1.0397\ne\ln2\approx0.6931$.

Remark. This is an instance of *Riemann's rearrangement theorem*: a conditionally convergent series can be rearranged to converge to any real number, or to diverge. Only absolutely convergent series are safe to rearrange.