Solve $\int _0^x \sin (t^2) \text{d}t$ using MacLaurin series (for $x \in \mathbb{R}$)

Question

We know that $\sin (t)=\sum _{n=0}^{\infty} (-1)^n \dfrac{t^{2n+1}}{(2n+1)!}$, and this implies that $\sin (t^2)=\sum _{n=0}^{\infty} (-1)^n \dfrac{(t^2)^{2n+1}}{(2n+1)!}=\sum _{n=0}^{\infty} (-1)^n \dfrac{t^{4n+2}}{(2n+1)!}$. Now, to solve the integral, I have to use the identity: $$\int _0^x \sin (t^2) \text{d}t=\int_0^x \sum_{n=0}^\infty(-1)^n\dfrac{t^{4n+2}}{(2n+1)!}\text{d}t=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}\int_0^xt^{4n+2}\text{d}t$$ The problem is that I can exchange integral and infinite sum only when the power series converges. In this case, I can rewrite my series into a power series the following way: $$\sum_{n=0}^\infty (-1)^n\dfrac{t^{4n+2}}{(2n+1)!}=\sum_{n=0}^\infty a_nt^n \qquad \text{where }a_{n}= \begin{cases}0 & \text { if } \not\exists k \in \mathbb{N}:n=4k+2 \\ \dfrac{(-1)^{(n-2)/4}}{\left(\frac{n}{2}\right)!} & \text { otherwise }\end{cases}$$ So we can calculate the radius of convergence $\rho$ by using d'Alembert theorem: $$\limsup _{n \rightarrow +\infty} \left|\dfrac{a_{n+1}}{a_n}\right|=?$$ We want $\rho =+\infty$, because the integration domain is $(0,x)$ with $x \in \mathbb{R}$, so we must have a power series that converges for every $t \in \mathbb{R}$, so the $\limsup$ must be $0$, but how can I calculate it formally? We have that the possibilities for $|a_{n+1}/a_n|$ are $0,0/0,(1/(n/2)!)/0$, so it seems to me that this $\limsup$ is $0$, but I don't know how to say it.