Bounding the $n$-th derivative of a periodic function

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Suppose I have a function $f(x) \in C^{\infty}$ that is periodic over $[0,2\pi]$. I want to find an upper bound on $$ \left | \frac{d^nf}{dx^n} \right |, $$

but I'm not sure how to do it.

I know that such a function can be expanded in a Fourier series:

$$ f(x) = \sum_{k = 0}^{\infty} a_k e^{ikx} $$

and that

$$ \left | \frac{d^nf}{dx^n} \right | = \left | \sum_{k = 0}^{\infty} (ik)^n a_k e^{ikx} \right | \leq \sum_{k=0}^{\infty} |a_k| k^n $$

I believe that asymptotically, $|a_k| = O(e^{-k})$ (from the convergence of the Fourier series), so we would get a sum like $\sum_{k}k^n e^{-k}$ but I'm not sure if this is a correct next step and, if so, how to bound this sum.