Prove $g(x) = \sum_{n=1}^{∞}\int_{0}^{\frac{x}{\sqrt{n}}} \sin t^2 dt$ is well-defined and differentialbe on $(0,∞)$

Prove $$g(x) = \sum_{n=1}^{∞}\int_{0}^{\frac{x}{\sqrt{n}}} \sin t^2 dt$$ is well-defined and differentialbe on $$(0,∞)$$

Well-defined:

To prove that $$g(x)$$ is indeed well-defined, it should suffice to take any $$x_0\in(0,∞)$$ and show that the series of integrals converge. Of course the necessary condition for convergence is met, as $$\int_{0}^{\frac{x_0}{\sqrt{n}}} \sin t^2 dt \rightarrow0 \quad \text{as} \quad n\rightarrow ∞.$$ For any $$x_0\in(0,∞)$$, thare exists $$n_0$$ s.t. the integral above will be strictly greater than $$0$$. So I could use a comparison test for series, but I have trouble finding $$a_n$$ to bound the integral from above: $$0 < \int_{0}^{\frac{x_0}{\sqrt{n}}} \sin t^2 dt < a_n \quad (\sum a_n \text{converges})$$

Differentiability:

Let $$f_n(x) = \int_{0}^{\frac{x}{\sqrt{n}}} \sin t^2 dt$$. Now if $$\sum ||f'_n(x)||_∞$$ converges, then I fould use the fact that $$\frac{d}{dx}\sum f_n = \sum \frac{d}{dx} f_n$$ But $$||f'_n(x)||_∞ = \sup_{x\in (0,∞)}\left(\frac{1}{\sqrt{n}}\cdot \sin(\frac{x^2}{n})\right) = \frac{1}{\sqrt{n}}$$, so $$\sum ||f'_n(x)||_∞$$ doesn't converge. I've ran out of ideas, I will appreciate any hint.

Hint: For both parts the inequality $$\sin y which is true for positive $$y$$ is useful. For small enough $$y$$ you even have $$0<\sin y. As for the convergence of the series of derivatives, it is a common trick to prove convergence on $$\left(0,M\right)$$ for every $$M>0$$.