= Paired logarithmic series with zero sum
{title2=$\sum_{n=2}^{\infty}\ln[1+(-1)^n/n]=0$}
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 $\ln[(2m+1)/(2m)]\to0$. 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 $\ln(1+u)\geq u/(1+u)$ for $u\geq0$, 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.
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