= Solution
The <asymptotic expansion> means that for every fixed integer $M\geq0$,
$$
f(n)-\sum_{k=0}^Ma_kn^{-2k}=o(n^{-2M})\qquad(n\to\infty).
$$
It concerns the successive finite truncations; the infinite series need not converge. Because the assertion holds at every order, one may equivalently bound the remainder after order $M$ by $O(n^{-2M-2})$.
Put $h(t)=(1+t^2)^{-1}$ and $b=2\pi$. For integer $n$, $\sin(nb)=\sin0=0$ and $\cos(nb)=\cos0=1$. Two <integrations by parts> give
$$
I(n)=\frac{h'(b)-h'(0)}{n^2}-\frac1{n^2}\int_0^bh''(t)\cos(nt)\,dt.
$$
Repeating exactly $M$ times gives
$$
I(n)=\sum_{k=1}^M\frac{(-1)^{k-1}[h^{(2k-1)}(b)-h^{(2k-1)}(0)]}{n^{2k}}+\frac{(-1)^M}{n^{2M}}\int_0^bh^{(2M)}(t)\cos(nt)\,dt.
$$
All derivatives are integrable on this compact interval. The <Riemann-Lebesgue lemma> says their oscillatory integrals tend to zero, proving the expansion at every order, including $a_0=0$. Since $h$ is even, its odd derivatives vanish at zero. Now $h'(t)=-2t/(1+t^2)^2$ and $h'''(t)=24t(1-t^2)/(1+t^2)^4$, so
$$
\boxed{a_0=0,\qquad a_1=-\frac{4\pi}{(1+4\pi^2)^2},\qquad a_2=\frac{48\pi(4\pi^2-1)}{(1+4\pi^2)^4}.}
$$
The restriction to integer $n$ is essential to this even-power endpoint expansion: otherwise the first integration by parts has a generally nonzero $n^{-1}$ boundary term.
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