= Solution
Let $a_n=\ln[1+(-1)^n/n]$ and $S_N=\sum_{n=2}^Na_n$. Consecutive even and odd terms cancel exactly:
$$
a_{2j}=\ln\frac{2j+1}{2j},\qquad
a_{2j+1}=\ln\frac{2j}{2j+1}=-a_{2j}.
$$
Thus
$$
S_{2m+1}=0,\qquad S_{2m}=\ln\frac{2m+1}{2m}\longrightarrow0.
$$
Both parity subsequences have the same limit, so the full sequence of <partial sums> converges and the sum is \b[zero]. This check is necessary: cancellation in fixed pairs alone does not always prove convergence of the original series.
For <absolute convergence>, each pair instead contributes
$$
|a_{2j}|+|a_{2j+1}|=2\ln\left(1+\frac1{2j}\right)
\geq\frac{2}{2j+1}.
$$
The inequality follows from $\ln(1+u)=\int_0^u(1+s)^{-1}ds\geq u/(1+u)$. The lower-bound series diverges by comparison with the <harmonic series>. Therefore the <paired logarithmic series with zero sum> has
$$
\boxed{\text{conditional convergence, but not absolute convergence}.}
$$
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