# $\sum_{n=1}^{\infty}\frac{\sin (nx^2)}{1+n^3}$ 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 ?

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 $$f'(x)=\sum_{n=1}^\infty f'_n(x),\quad x\in\mathbb{R}.$$
• Well..You show that $\sum f_n'$ is uniformly convergent on $[-R,R]$..but not in $\mathbb R$. My question is whole over $\mathbb R$. From uniform convergence of $\sum f_n'$ on $[-R,R]$ how we can conclude the result ? – user181525 Mar 24 '15 at 14:54
• $f$ id differentiable on $[-R,R]$ for all $R>0$, thus it is differentiable on $(-\infty,\infty)$. – Julián Aguirre Mar 24 '15 at 16:01
• $|f_n'(x)|\le \frac{2R}{n^2}$. If $R$ is finite then $\sum f_n'$ is uniformly convergent...But when $R\to \infty$ then how you conclude that $\sum f_n'$ is uniformly convergent ? – user181525 Mar 24 '15 at 16:29
• I do not conclude that. My conclusion is that $\sum f'_n$ converges to $f'$ uniformly on compact intervals and pointwise on $\mathbb{R}$. – Julián Aguirre Mar 24 '15 at 18:07
• So the condition that a given series represents a differentiable function that : " $\sum f_n$ is convergent and $\sum f_n'$ is uniformly convergent on any bounded interval ". Is it ? – user181525 Mar 24 '15 at 18:56