Has the fourier series for all smooth periodic functions finitely many terms?

Question

When a fourier series has only finitely many terms, $f(x) := \sum_{n=-N}^N f_n e^{inx}$, then it is obvious that $f$ is smooth and periodic. Is the converse true aswell?

If we have a smooth periodic function, can we then conclude that the fourier series has finitely many terms?

There is a corollary that for a $2 \pi$-periodic function $f\in C^k(\mathbb{R}/2\pi)$ the fourier coefficients $f_n$ we have $|n|^k|f_n| \rightarrow 0$ as $k \rightarrow \infty$ but i am not sure how that helps and i can not think of any counter example.

Answer 1

$f(x)=\sum_{-\infty}^{\infty} \frac {e^{inx}} {n!}$ is periodic, infinitely differentiable and has infinite number of terms.

Answer 2

The simplest approach is to define a function that we know takes infinitely many terms. Take your favorite converging infinite series and make the terms the $f_n$. Mine is $ \frac 1{n^2}$ giving $$f(x)=\sum_{n=-\infty}^{\infty}\frac {e^{inx}}{n^2}$$For a given $x$ the series is absolutely convergent.