= Solution
Let $a_n$ denote the summand. Its denominator is positive for every $n\ge1$: odd $n$ gives $n^4+n^2$, while even $n\ge2$ gives $n^4-n^2$. Dividing numerator and denominator by their highest powers gives
$$
na_n=\frac{1+8(-1)^n/n+1/n^3}{1+(-1)^{n+1}/n^2}\longrightarrow1.
$$
The terms are positive for sufficiently large $n$, so the <limit comparison test> with the divergent <harmonic series> gives \b[divergence to $+\infty$]. More explicitly, for $n\ge16$ the numerator is at least $n^3/2$ and the denominator is at most $2n^4$, giving $a_n\ge1/(4n)$. Any finitely many early negative terms change partial sums by only a finite constant and cannot prevent divergence of this positive tail.
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