Why is $\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ continuously differentiable in $\mathbb{R}$?

Question

Let $M>0$. In the closed interval $[-M,M]$ we get that:

1. $\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ converges at $x_{0}=0$
2. $\sum_{n=1}^{\infty}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'=\sum_{n=1}^{\infty}\frac{2nx\cos(nx^{2})}{1+n^{4}}$ and $\left|\frac{2nx\cos(nx^{2})}{1+n^{4}}\right|\leq\frac{2nM}{1+n^{4}}$. Also $\sum_{n=1}^{\infty}\frac{2nM}{1+n^{4}}<\infty$ so by the Weierstrass M-test, $\sum_{n=1}^{\infty}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'$ converges uniformely in $[-M,M]$.

Therefore, by the term by term differentiation theorem there exists a function $f\colon[-M,M]\to\mathbb{R}$ such that $f(x)=\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ uniformly and $\left(\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}\right)'=f'(x)=\sum_{n=1}^{\infty}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'$.

Finally, $\sum_{n=1}^{N}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'$ is continuous in $[-M,M]$ for every $N$, then by the uniform convergence $f'(x)=\sum_{n=1}^{\infty}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'$ is also continuous in $[-M,M]$.

My question is how can i generalize this argument to $\mathbb{R}$?

Answer 1

Apparently what i was missing is this:

Let $x_{0}\in\mathbb{R}$ ,

We need to show that $\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ is continuously differentiable at $x_{0}$. That is to show that $\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ is differentiable at $x_{0}$ and that $\left(\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}\right)'$ is continuous at $x_{0}$.

Choose $M=\left|x_{0}\right|+1$ so $x_{0}\in[-M,M]$ and from the argument above we get what we need.