I'm trying to prove: for $f \in C^1[0, 2\pi]$ such that $f(0)=f(2\pi)$ show that $\sum_{k=-\infty}^{\infty}\vert \langle f, e_k \rangle \vert < \infty$, where $e_k(t)=e^{-ikt}$ for $k \in \mathbb{Z}$.
So I have: $\sum_{k=-\infty}^{\infty}\vert \langle f, e_k \rangle \vert = \sum_{k=-\infty}^{\infty}\vert \int_0^{2\pi} f(t) \cdot e^{-ikt}dt \vert =\ ... = \sum_{k=-\infty}^{\infty} \vert \int_0^{2\pi}k f'(t) \cdot e^{-ikt}dt \vert= \sum_{k=-\infty}^{\infty}\vert k \ \langle f', e_k \rangle \vert $.
I have problems with the next step. How do I show the statement?
For $k \in \mathbb{Z}$ and $f \in C^1([0, 2\pi])$ we have:
$\langle f, e_k \rangle = \frac{1}{\sqrt{2\pi}}\int_0^{2\pi}f(t)e^{-ikt}dt= \frac{1}{ik}\langle f', e_k \rangle$ (As you can see I was wrong about this term). Then we have for $k \in \mathbb{N}$:
$\sum_{k=1}^K \vert \langle f, e_k \rangle \vert = \sum_{k=1}^K \frac{1}{k} \vert \langle f', e_k \rangle \vert \leq \Big{(}\sum_{k=1}^K \frac{1}{2k^2}\Big{)}^{\frac{1}{2}} \Big{(} \sum_{k=1}^K \vert \langle f', e_k \rangle \vert^2 \Big{)}^{\frac{1}{2}} \leq (\zeta(2))^{\frac{1}{2}} \Big{(} \sum_{K \geq 1} \vert \langle f', e_k \rangle \vert^2 \Big{)}^{\frac{1}{2}} < \infty$,
hence $\sum_{k=1}^{\infty} \vert \langle f, e_k \rangle \vert < \infty$. The same proof for $\sum_{k=1}^{\infty} \vert \langle f, e_{-k} \rangle \vert < \infty$.
So we have $\sum_{k \in \mathbb{Z}} \vert \langle f, e_k \rangle \vert < \infty \ $ q.e.d