$\sum_{n=1}^{\infty}\frac{\sin (nx^2)}{1+n^3}$ represents a differentiable function Show that the following series of function defines a continuous differentiable function function in $\mathbb R$.
$$\sum_{n=1}^{\infty}\frac{\sin (nx^2)}{1+n^3}.$$
We have , $|f_n(x)|=\left|\frac{\sin (nx^2)}{1+n^3}\right|\le \frac{1}{1+n^3}\le \frac{1}{n^3}=M_n \text{ (say) }$
As, $\sum M_n$ is convergent so, the given series is uniformly convergent. Also , each $f_n(x)$ is a continuous function in $\mathbb R$. So, the given series converges to a continuous function , say $f(x)$.
But how we can show that $f(x)$ is differentiable function ?
Please help...

Thew derivative of $f_n$ is continuous and is given by
$$
f'_n(x)=\frac{2\,n\,x\sin (n\,x^2)}{1+n^3}.
$$
If $|x|\le R$, then
$$
|f'_n(x)|\le\frac{2\,n\,R}{1+n^3}\le\frac{2\,R}{n^2}.
$$
Since $\sum1/n^2<\infty$, The series $\sum f'_n$ is uniformly convergent on $[-R,R]$. This, together with the convergence of $\sum f_n$, proves that $f$ is differentiable and 
$$
f'(x)=\sum_{n=1}^\infty f'_n(x),\quad x\in\mathbb{R}.
$$
