Here's the theorem:
Theorem: If $f$ is periodic with Fourier coefficients $a_n,b_n$ and if the series $$\sum_{n=1}^\infty (|n^{k}a_n|+|n^{k}b_n|)$$ converges for some integer $k \geq 1$, then f has continuous derivatives $f',...,f^{(k)}$ whose Fourier series are differentiated series of $f$. $\blacksquare$
So we know that $f(x)=\sum_{n=0}^\infty a_ncos(nx)+b_nsin(nx)$ on the interval $[-\pi,\pi]$. What needs to be done next is proving that the derivative of our series $f$ converges uniformly on our interval. I know how to prove the inequality $$\sum_{n=1}^\infty [a_n(cos(nx))^{(h)}+b_n(sin(nx))^{(h)}] \leq \sum_{n=1}^\infty (|n^{h}a_n|+|n^{h}b_n|) \leq \sum_{n=1}^\infty (|n^{k}a_n|+|n^{k}b_n|)$$ for all $ h \in [1,k]$ via the Weierstrass M-Test. Using induction, is it sufficient to show $f'$ is uniformly convergent and work my way up from there to the $f^{(k)}$ case, where assume the convergence of the $(k-1)^{th}$ derivative of our series to be equal to $f^{(k-1)}$? Will that ultimately show that $$f^{(k)}(x)=\sum_{n=1}^\infty [a_n(cos(nx))^{(k)}+b_n(sin(nx))^{(k)}]?$$
If this is the case (which I'm almost positive it is based on the way uniform convergence is defined), then here is my attempt at a proof of this. If anyone could tell me if I did this even remotely correctly, I would quite grateful!
Attempted proof: Consider $$f(x)=a_0 + \sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)]$$ which is $2\pi$-periodic. Differentiating $f$, we obtain $$f'(x) \sim \frac{d}{dx}(\sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)]).$$ Recalling that $a_n=\int_{-\pi}^{\pi}f(x)cos(nx)dx$ and $b_n=\int_{-\pi}^{\pi}f(x)sin(nx)dx$, it using integration by parts that $$a_n'=\int_{-\pi}^{\pi}f'(x)cos(nx)dx=nb_n$$ and $$b_n'=\int_{-\pi}^{\pi}f'(x)sin(nx)dx=-na_n$$ $$\implies f'(x) \sim \frac{d}{dx}(\sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)])$$ $$=\sum_{n=1}^\infty [-na_nsin(nx)+nb_ncos(nx)]$$ $$\equiv \sum_{n=1}^\infty [a_n\frac{d}{dx}(cos(nx))+b_n\frac{d}{dx}(sin(nx))].$$ Let's show that $\sum_{n=1}^\infty [-na_nsin(nx)+nb_ncos(nx)]$ is uniformly convergent on $[-\pi,\pi]$, which will show that that our differentiated Fourier series converges to $f'$ on $[-\pi,\pi]$. Recall that $|cos(x)| \leq 1$ which extends to $|cos(nx)| \leq 1$ and $|sin(x)| \leq 1$ which extends to $|sin(nx)| \leq 1$. Thus, $$\sum_{n=1}^\infty [-na_nsin(nx)+nb_ncos(nx)]$$ $$\leq |\sum_{n=1}^\infty [-na_nsin(nx)+nb_ncos(nx)]|$$ $$\leq \sum_{n=1}^\infty |-na_nsin(nx)+nb_ncos(nx)|$$ $$\leq \sum_{n=1}^\infty [|-na_nsin(nx)|+|nb_ncos(nx)|]$$ $$\leq \sum_{n=1}^\infty (|na_n|+|nb_n|)$$ $$ \equiv \sum_{n=0}^\infty (|n^1a_n|+|n^1b_n|)]$$ so we see our series is uniformly convergent on $[-\pi,\pi]$ by the Weierstrass M-Test, which implies that $f'$ represents our series on $[-\pi,\pi]$ and we have proved the case of $k=1$.
Now suppose $$f^{(k-1)}(x)=\frac{d^{(k-1)}}{dx^{(k-1)}}(\sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)])$$ $$\equiv \sum_{n=1}^\infty [a_n\frac{d^{(k-1)}}{dx^{(k-1)}}(cos(nx))+b_n\frac{d^{(k-1)}}{dx^{(k-1)}}(sin(nx))]$$ on the interval $[-\pi,\pi]$. We have currently have, since we don't know if the k-th derivative of $f$'s Fourier series converges to $f^{(k)}$, $$f^{(k)} \sim \sum_{n=1}^\infty [a_n\frac{d^{(k)}}{dx^{(k)}}(cos(nx))+b_n\frac{d^{(k)}}{dx^{(k)}}(sin(nx))].$$ This part of the proof is actually proven in an older post on stackexchange, which I link here:
It follows, finally, that $$f^{(k)}=\sum_{n=1}^\infty [a_n\frac{d^{(k)}}{dx^{(k)}}(cos(nx))+b_n\frac{d^{(k)}}{dx^{(k)}}(sin(nx))]$$ on the interval $[-\pi,\pi]$, so by the Principle of Mathematical Induction, we see that $f$ has $k$-continuous derivatives given the conditions imposed. $$\blacksquare$$
