Okay so here's the the problem:
Let $k \in \mathbb{Z}.$ Then, $\forall$ $m \in [1,k]$, $$f^{(m)}(x) = \sum_{n=1}^\infty{[a_ncos(nx)]^{(m)}+[b_nsin(nx)]^{(m)}} \leq \sum_{n=1}^\infty{(|a_nn^m| + |b_nn^m|)}$$
Now I'm not sure really sure how to approach this. A friend of mine asked me for some aid with this, saying to use the "Differentiation Theorem of a function" from Rudin's text. I really wasn't sure which Differentiation Theorem he was speaking of and trying to dig deeper, he said that's all he knew on the way to approach this. I tried looking at Lebesgue's Differentiation Theorem but it seemed far too complicated for this problem. My friend supplied me with a Theorem and its corollary that lead to the proposition:
Theorem: If $f$ is a periodic function with Fourier coefficients $a_n,b_n$ and the series $\sum_{n=1}^\infty(|a_nn^k|+|b_nn^k|)$ converges for some $k \in \mathbb{N}$, then $f$ has continuous derivatives $f',f'',...,f^{(m)}$
Corollary: Let $k \in \mathbb{N}$. If $f$ is periodic, with Fourier coefficients $a_n,b_n$ and the series $\sum_{n=1}^\infty{(|a_n| + |b_n|)|n^k|}$ converges for some $k$, then $\forall$ $m \in [0,k]$ it follows that $$f^{(m)}(x)=\frac{d^m}{dx^m}[a_0+\sum_{n=1}^\infty({a_ncos(nx)+b_nsin(nx)})] $$
The proposition clearly sets up an inequality between the Theorem and its Corollary. I'm just quite unsure where to go with this. If anyone could give a hand, my friend and I would be much obliged.
EDIT****EDIT****EDIT
Thanks to Michael Albanese, I think I've got the answer now. Here is my attempt at a formal proof
Proposition: Prove $\forall$ $m \in [1,k]$, where $k \in \mathbb{N}$, that $$\frac{d^m}{dx^m}(\sum_{n=1}^\infty(a_ncos(nx)+b_nsin(nx))) \leq \sum_{n=1}^\infty(|a_nn^m|+|b_nn^m|)$$
Proof
Consider $\mathbb{Z_4}$, where we have equivalence classes $[0]_4,[1]_4,[2]_4,[3]_4$. Now observe that $\forall$ $m \in [1,k]$ $$\frac{d^m}{dx^m}(cos(nx))=\left\{ \begin{array}{lr} n^mcos(nx) & m\equiv [0]_4\\ -n^msin(nx) & m\equiv [1]_4\\ -n^mcos(nx) &m\equiv [2]_4\\ n^msin(nx) &m\equiv [3]_4 \end{array} \right.$$
$$\frac{d^m}{dx^m}(sin(nx))=\left\{ \begin{array}{lr} n^msin(nx) & m\equiv [0]_4\\ n^mcos(nx) & m\equiv [1]_4\\ -n^msin(nx) &m\equiv [2]_4\\ -n^mcos(nx) &m\equiv [3]_4 \end{array} \right.$$
Further,
$$\frac{d^m}{dx^m}(cos(nx)) \leq \bigl|\frac{d^m}{dx^m}(cos(nx))\bigr|=\left\{ \begin{array}{lr} |n^mcos(nx)| & m\equiv [0]_4\\ |n^msin(nx)| & m\equiv [1]_4\\ |n^mcos(nx)| &m\equiv [2]_4\\ |n^msin(nx)| &m\equiv [3]_4 \end{array} \right. $$ $$\frac{d^m}{dx^m}(sin(nx)) \leq \bigl|\frac{d^m}{dx^m}(sin(nx))\bigr|=\left\{ \begin{array}{lr} |n^msin(nx)| & m\equiv [0]_4\\ |n^mcos(nx)| & m\equiv [1]_4\\ |n^msin(nx)| &m\equiv [2]_4\\ |n^mcos(nx)| &m\equiv [3]_4 \end{array} \right.$$
Recall now that $\bigl|cos(x)\bigr| \leq 1$ and $\bigl|sin(x)\bigr| \leq 1$ which extends to $nx$. Thus,
$$\bigl|\frac{d^m}{dx^m}(cos(nx))\bigr| \leq \left\{ \begin{array}{lr} |n^m| & m\equiv [0]_4\\ |n^m| & m\equiv [1]_4\\ |n^m| &m\equiv [2]_4\\ |n^m| &m\equiv [3]_4 \end{array} \right. \equiv |n^m|$$ $$\bigl|\frac{d^m}{dx^m}(sin(nx))\bigr| \leq \left\{ \begin{array}{lr} |n^m| & m\equiv [0]_4\\ |n^m| & m\equiv [1]_4\\ |n^m| &m\equiv [2]_4\\ |n^m| &m\equiv [3]_4 \end{array} \right. \equiv |n^m|$$
since for both $\bigl|\frac{d^m}{dx^m}(cos(nx))\bigr|$ and $\bigl|\frac{d^m}{dx^m}(sin(nx))\bigr|$, we see our last inequalities result in $|n^m|$ for $[0]_4,[1]_4,[2]_4,[3]_4$. Therefore, it follows that
$$\frac{d^m}{dx^m}(\sum_{n=1}^\infty(a_ncos(nx)+b_nsin(nx))) = \sum_{n=1}^\infty(a_n\frac{d^m}{dx^m}((cos(nx))+b_n\frac{d^m}{dx^m}((sin(nx)))$$ $$\leq \bigl|\sum_{n=1}^\infty (a_n\frac{d^m}{dx^m}((cos(nx))+b_n\frac{d^m}{dx^m}((sin(nx))\bigr)|$$ $$\leq \sum_{n=1}^\infty (\bigl|a_n\frac{d^m}{dx^m}((cos(nx))+|b_n\frac{d^m}{dx^m}((sin(nx))\bigr|)$$
$$\leq \sum_{n=1}^\infty (\bigl|a_n\frac{d^m}{dx^m}((cos(nx))\bigr|+\bigl|b_n\frac{d^m}{dx^m}((sin(nx))\bigr|)$$ $$\leq \sum_{n=1}^\infty(|a_nn^m|+|b_nn^m|)$$
$$\implies \frac{d^m}{dx^m}(\sum_{n=1}^\infty(a_ncos(nx)+b_nsin(nx))) \leq \sum_{n=1}^\infty(|a_nn^m|+|b_nn^m|)$$ $$\blacksquare$$
