= Solution
For $s=\pm1$, the identity $s/(2+s)=-1/3+(2/3)s$ gives
$$
\frac{(-1)^n}{n(2+(-1)^n)}
=-\frac1{3n}+\frac23\frac{(-1)^n}{n}.
$$
The <alternating series test> applies to $\sum(-1)^n/n$ because $1/n$ decreases to zero; its <partial sums> consequently have a finite limit. The <harmonic series> $\sum1/n$ diverges to $+\infty$. Applying the displayed identity to each finite <partial sum> yields
$$
\sum_{n=1}^N\frac{(-1)^n}{n(2+(-1)^n)}
=-\frac13\sum_{n=1}^N\frac1n
+\frac23\sum_{n=1}^N\frac{(-1)^n}{n}
\longrightarrow-\infty.
$$
Therefore $\boxed{\text{the series diverges to }-\infty}$. Alternating signs alone do not satisfy the <alternating series test>: the magnitudes jump from $1/(3n)$ at even $n$ to $1/(n+1)$ at the next odd index.
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