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?
Best Answer
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
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.