 Solution

The asymptotic expansion means that for every fixed integer $M\ge 0$,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 $n$, $\sin (nb)=\sin 0=0$ and $\cos (nb)=\cos 0=1$. Two integrations by parts giveRepeating exactly $M$ times givesAll 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 $h^{\prime }(t)=-2t/(1+t^2{)}^2$ and $h^{\prime \prime \prime }(t)=24t(1-t^2)/(1+t^2{)}^4$, soThe 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.