Find the exact number of times $f(x)=\displaystyle\sum_{n\neq 0}\frac{1}{n^{4.01}}e^{inx}$ is continuously differentiable. (find the greatest $k\in\mathbb{N}$ such that $f\in C^k$)
According to theorem $(1)$ - if $f$ is a $2\pi$ periodic function such that $\displaystyle\sum_{n=-\infty}^{\infty}|n^k\hat f(n)e^{inx}| < \infty$, that implies $f\in C^k$. According to theorem (2)- if $f\in C^k$ (and $f$ is $2\pi$ periodic) then $\lim_{n\to±\infty}|n^k \hat f(n)|= 0$.
Since $\lim_{n\to±\infty}|n^5 \frac{1}{n^{4.01}}| \neq 0 \implies f\notin C^5$,
and since $\displaystyle\sum_{n=-\infty}^{\infty}|n^3. \frac{1}{n^{4.01}}|$ converges, then $f\in C^3$. Hence $k=3$ is the minimum and $k=4$ could be the maximum. I need to determine if $k=3$ or $k=4$, yet I have no idea how to continue from here. Help would be appreciated, thanks in advance :)