= Solution
Rationalize the positive difference:
$$
\sqrt{n^4+a^2}-n^2=\frac{a^2}{\sqrt{n^4+a^2}+n^2}.
$$
Since the denominator is at least $2n^2$,
$$
0<\sqrt{n^4+a^2}-n^2\le\frac{a^2}{2n^2}.
$$
The upper bound is a constant multiple of the convergent <p-series> with exponent two. The <comparison test> therefore proves
$$
\boxed{\sum_{n=1}^\infty\left(\sqrt{n^4+a^2}-n^2\right)\text{ converges for every fixed }a>0.}
$$
The <asymptotic equivalence> of its summand to $a^2/(2n^2)$ gives the same conclusion by <limit comparison test>.
Back to article page
