Define the formal series

f(x) = \sum_{n \geq 1} \frac{(-1)^n}{n^3} \sin(2\pi n x), \text{ let } a < b.

Let $S_n(x) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \sin(2\pi n x)$ , for $n \geq 1$ .

Note that $S_n$ is continuous and differentiable in $\mathbb{R}$ , for being a finite sum of such functions.

Moreover

S_n^{(x)} = 2\pi \sum_{n \geq 1}^{\infty} \frac{(-1)^n}{n^2} \cos(2\pi n x) \in \mathcal{C}[a, b].

Note that for all $n \geq 1$ and $x \in \mathbb{R}$

\left| \frac{(-1)^n a \geq (2\pi n x)}{n^2} \right| + \left| \frac{(-1)^n}{n^3} \sin(2\pi n x) \right| \leq \frac{1}{n^2}

and the series $\sum_{n \geq 1} \frac{1}{n^2}$ converges.

By the $\pi$ -test of Weierstrass

& $S_n^(\Rightarrow I) \xrightarrow{(-1)^n} \cos(2\pi n x) = g$

& $S_n \Rightarrow f \quad \text{as } N \to \infty$ .

Hence, we have that $f$ is differentiable in $(a, b)$ and $f'(x) = I \xrightarrow{(-1)^n} \cos(2\pi n x)$ .

For all $x \in (a, b)$ , and $g \in \mathcal{C}[a, b]$ .

Since $a, b$ are arbitrary, we conclude that $f \in \mathcal{C}^1(\mathbb{R})$ and $f' \in \mathcal{C}(\mathbb{R})$