# Is the sum of the series $\sum \frac{\sin nx^2}{1 + n^3}$ continuously differentiable?

Consider series $$\sum_{n=1}^\infty \frac{\sin nx^2}{1 + n^3}$$ Is its limit continuously differentiable?

We have for all $x \in \mathbb{R}$, $$\left|\frac{\sin nx^2}{1 + n^3}\right| \leq \frac{1}{1 + n^3} < \frac{1}{n^3}$$ By the Weierstrass M-test, as $\sum \frac{1}{n^3}$ is convergent, our series is uniformly convergent. Thus, its limit function is continuous.

To show that its limit is continuously differentiable, if $f_n(x) = \frac{\sin nx^2}{1 + n^3}$, it suffices to show that $\sum f_n'$ converges uniformly on $\mathbb{R}$, where $$f_n'(x) = \frac{2nx\cos nx^2 }{1 + n^3}$$ (As $f_n'$ is continuous for each $n \in \mathbb{N}$, uniform convergence would also imply that the derivative is continuous).

However, I don't know where this series converges uniformly or not. We do have $$\left|\frac{2nx\cos nx^2 }{1 + n^3}\right| \leq \frac{2n|x|}{1 + n^3}$$ By the comparison test, we can conclude this series converges pointwise. Does it converge uniformly? If not, how do we prove whether or not the original series is continuously differentiable?

You can consider a compact set $K=\left[a,b\right]$ with $\left(a,b\right) \in \mathbb{R}^{2}$ and $a<b$. The function $\displaystyle x \mapsto \frac{2nx\cos\left(nx^2\right)}{1+n^3}$ is odd and then we can study it only on $\mathbb{R}^{+}$. For $x \in K$ you now have $$\left|\frac{2nx\cos\left(nx^2\right)}{1+n^3}\right| \leq \frac{2nb}{1+n^3}$$ Hence
$$\left\|\frac{2nx\cos\left(nx^2\right)}{1+n^3}\right\|_{\infty,K} \leq \frac{2nb}{1+n^3}\underset{(+\infty)}{\sim}\frac{2b}{n^2}$$
The series $\displaystyle \sum_{n \geq 1}^{ }\frac{1}{n^2}$ converges hence the series $\displaystyle \sum_{n \geq 0}^{ }f'_n$ converges normally so also uniformly on every compact set of $\mathbb{R}$ ( remember it was odd ). Then it is differentiable on $\mathbb{R}$ because it is a local property.
• Great, I didn't realize that it is enough to show uniform convergence on any bounded interval, but I see now that it is. Out of curiosity, is $\sum f_n'$ uniformly convergent on $\mathbb{R}$? Or indeed, is there an easy way to answer that? – Anu Jan 31 '18 at 12:30
• I cannot tell you like that but my feeling is that it is not because you will have difficulties majoring $x$ on $\mathbb{R}^{+}$. But I might be wrong because I have not proved it. – Atmos Jan 31 '18 at 12:33