= Solution
The positive terms $b_n=1/\log(n+1)$ decrease to zero. The <alternating series test> just proved therefore gives \b[convergence] of the signed <series>. It is <conditional convergence>: $\log(n+1)\le n$, so $b_n\ge1/n$, and the absolute-value <series> dominates the divergent <harmonic series>.
The <comparison test for series> used here says that if $0\le c_n\le d_n$, convergence of $\sum d_n$ implies convergence of $\sum c_n$, while divergence of $\sum c_n$ implies divergence of $\sum d_n$. This follows by comparing increasing <partial sums>. For completeness, the <harmonic series> diverges because every block from $2^k$ to $2^{k+1}-1$ has sum at least $1/2$.
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