Fresnel series geometric tail bounds

If $z=0$, all terms in [F1] vanish, and their division-free bounds and zero tails hold directly. For $z\mathrm{\ne }0$, every term is nonzero, and taking ratios in [F1] gives $\frac{\mathrm{\mid }{c}_{k+1}\mathrm{\mid }}{\mathrm{\mid }{c}_k\mathrm{\mid }}=\frac{{\pi }^2\mathrm{\mid }z{\mathrm{\mid }}^4}{4}\frac{4k+1}{(2k+1)(2k+2)(4k+5)},\frac{\mathrm{\mid }{s}_{k+1}\mathrm{\mid }}{\mathrm{\mid }{s}_k\mathrm{\mid }}=\frac{{\pi }^2\mathrm{\mid }z{\mathrm{\mid }}^4}{4}\frac{4k+3}{(2k+2)(2k+3)(4k+7)}\mathrm{.}$ The omitted factors are each less than $1$, yielding the displayed bounds.

The denominators in the bounds grow quadratically in $k$, so the ratios tend to zero; summing the resulting geometric majorant gives the stated tail bound and uniform convergence on every disk.

The infinite-radius power series in [F1] may be differentiated termwise by [F3]. The resulting derivatives are $C^{\mathrm{\prime }}(z)={\sum }_{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k(\pi \mathrm{/}2{)}^{2k}{z}^{4k}}{(2k)!}=\cos (\pi z^2\mathrm{/}2),S^{\mathrm{\prime }}(z)={\sum }_{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k(\pi \mathrm{/}2{)}^{2k+1}{z}^{4k+2}}{(2k+1)!}=\sin (\pi z^2\mathrm{/}2),$ where the last equalities use [F2].

Both $C$ and $S$ are entire primitives of their displayed integrands. For any piecewise-$C^1$ path $\gamma :[0,1]\to \mathbb{C}$ from $0$ to $z$, put ${\eta }_h=\gamma (s+h)-\gamma (s)$ on each smooth piece. Complex differentiability gives $C(\gamma (s)+{\eta }_h)-C(\gamma (s))=C^{\mathrm{\prime }}(\gamma (s)){\eta }_h+{\eta }_h{ϵ}_C({\eta }_h)$ with ${ϵ}_C(\eta )\to 0$ as $\eta \to 0$; set ${ϵ}_C(0)=0$. Since ${\eta }_h\mathrm{/}h\to {\gamma }^{\mathrm{\prime }}(s)$, division by $h$ makes the remainder tend to zero and proves $(C\circ \gamma {)}^{\mathrm{\prime }}(s)=C^{\mathrm{\prime }}(\gamma (s)){\gamma }^{\mathrm{\prime }}(s)$. The same argument gives $(S\circ \gamma {)}^{\mathrm{\prime }}(s)=S^{\mathrm{\prime }}(\gamma (s)){\gamma }^{\mathrm{\prime }}(s)$. At each piece endpoint the same limit with one-sided increments gives the relative derivatives; they extend continuously to the closed piece. By [F4], the real and imaginary parts integrate to endpoint differences. By [F6], the parameter integrals sum to the complex line integrals defined in [F5]. Since $C(0)=S(0)=0$, the line integrals equal $C(z)$ and $S(z)$. Hence they are path-independent and agree with the straight-segment integrals.