= Solution
The <limit comparison test> says that if $u_n,v_n>0$ and $u_n/v_n\to L$ with $0<L<\infty$, then their <series> either both converge or both diverge. This follows because, eventually, $(L/2)v_n\leq u_n\leq(3L/2)v_n$, and the <comparison test for series> applies in both directions.
Here comparison with the <harmonic series> gives
$$
\frac{1/[n+(\log n)^2]}{1/n}
=\frac1{1+(\log n)^2/n}\longrightarrow1.
$$
For completeness, the <natural logarithm> grows slowly enough because for $x\geq1$,
$$
0\leq\log x=\int_1^x\frac{dt}{t}
\leq\int_1^x t^{-3/4}\,dt
=4(x^{1/4}-1)\leq4x^{1/4},
$$
so $(\log n)^2/n\leq16/\sqrt n\to0$. The <harmonic series> diverges: every block from $n=2^k$ to $2^{k+1}-1$ has $2^k$ terms at least $2^{-k-1}$ and therefore contributes at least $1/2$. Hence
$$
\boxed{\sum_{n\geq1}\frac1{n+(\log n)^2}\text{ diverges to }+\infty.}
$$
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