will this function be Continuously differentiable? $x\in \mathbb{R}$ $$ f(x) = \sum_{n=1}^{\infty} {{\sin(nx^2)}\over 1+n^3} $$ will this be continuously differentiable? I know that Uniform convergence does not confirm the fact. That's why though the aforesaid series is convergent we can not ensure the fact that $f(x)$ will be continuously differentiable. Can anyone help me out by giving a rigorous proof?

If $f(x) = \sum_{n=1}^{\infty} {{\sin(nx^2)}\over 1+n^3}$, then $f(x)$ converges absolutely for all $x\in \mathbb{R}$ by the comparison test, as $|\frac{sin(nx^2)} {1+n^3}|\leq \frac{1}{1+n^3}$. And also $f'(x)=\sum_{n=1}^{\infty}{{\ 2nx \cos(nx^2)}\over 1+n^3}$. This is because for all $n \in \mathbb{N}$ the derivative of $f_m(x) = \sum_{n=1}^{m} {{\sin(nx^2)}\over 1+n^3}$ is $f_m'(x)=\sum_{n=1}^{m}{{\ 2nx \cos(nx^2)}\over 1+n^3}$ And since $|\frac{2nx \cos(nx^2)}{1+n^3}| \leq |\frac{2xn}{1+n^3}| \leq |\frac{2x}{n^2}$| and $\sum_{n=1}^{\infty}|{{2x}\over n^2}|$ converges, it follows from the Weierstrass M-test that the sequence $(f'_m)$ converges absolutely uniformly on any bounded interval and $\lim_{n\to\infty} f_n'(x)=f'(x)$. Thus, by the uniform limit theorem it follows that $f'(x)=\lim_{n\to\infty} f_n'(x)$ is continuous, and thus it follows that $f(x)$ is continuously differentiable.