 Paired logarithmic series with zero sum

The terms at $n=2j$ and $n=2j+1$ are $\ln [(2j+1)/(2j)]$ and its negative. Thus the partial sums ending at an odd index are zero, while those ending at $2m$ equal $\ln [(2m+1)/(2m)]\to 0$. Considering both subsequences proves convergence of the full sequence of partial sums, not merely of grouped terms.

The paired sum of absolute values is $2\ln [1+1/(2j)]$. Since $\ln (1+u)\ge u/(1+u)$ for $u\ge 0$, these pairs dominate $2/(2j+1)$, whose series diverges by comparison with the harmonic series. The original series therefore has conditional convergence with the displayed sum.