= Solution
Rationalize the difference to reveal its small size instead of comparing its two large summands separately:
$$
0\leq\sqrt{n^4+\alpha^4}-n^2=\frac{\alpha^4}{\sqrt{n^4+\alpha^4}+n^2}\leq\frac{\alpha^4}{2n^2}.
$$
For any fixed real $\alpha$, the right-hand <series> is a constant multiple of a convergent <P-series>. The <comparison test for series> proves
$$
\boxed{\text{convergence for every }\alpha\in\mathbb R.}
$$
The case $\alpha=0$ is the identically zero <series>. For nonzero $\alpha$, its summand is asymptotic to $\alpha^4/(2n^2)$, explaining the same convergence behavior as the inverse-square <series>.
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