What is this series? $\cos\frac{\pi z^2}{2}\sum_{n=1}^{\infty} \frac{\left(-1\right)^n \pi^{2n +1} z^{4n +3} }{1 \cdot 3 \cdots\left(4n + 3\right)}$

Since $~(4n+3)!!~=~\dfrac{(4n+3)!}{(4n+2)!!}~=~\dfrac{(4n+3)!}{(2n+1)!}\cdot\dfrac1{2^{2n+1}}~$ and $~\displaystyle\sum_{n=0}^\infty\frac{(2n+1)!}{(4n+3)!}~(2x)^{4n+3}=\dfrac{\sqrt\pi}2\cdot$

$\cdot\Big(e^{x^2}~\text{erf }x-e^{-x^2}~\text{erfi }x\Big),~$ it follows from Euler's formula that our infinite series, were it to

begin at $n=0$, would equal $~C(z)~\sin\bigg(\dfrac\pi2~z^2\bigg)-S(z)~\cos\bigg(\dfrac\pi2~z^2\bigg),~$ where C and S are the

two Fresnel integrals.