Let $\{a_k\}_{k=1}^\infty \subset \mathbb R$ be a sequence, and suppose there exists constant $M>0$ s.t. $|a_k|\leqq \dfrac{M}{k^3}$ for all $k.$
Then, prove that the series of function $\{f_n : \mathbb R \to \mathbb R\}_{n=1}^\infty$ where $f_n(x):=\displaystyle\sum_{k=1}^n a_k \sin (kx)$ uniformly converges to a $C^1$ function $g$ defined on $\mathbb R.$
My attempt is below.
Let $h_n(x):=f'_n (x)= \displaystyle\sum_{k=1}^n k a_k \cos kx,$ and $h(x):=\displaystyle\sum_{k=1}^\infty k a_k \cos kx.$ $\left( h(x) \mathrm{ \ exists \ since\ } \left|\displaystyle\sum_{k=1}^n k a_k \cos kx\right|\leqq \displaystyle\sum_{k=1}^n \dfrac{M}{k^2} \to \displaystyle\sum_{k=1}^\infty \dfrac{M}{k^2}<\infty.\right)$
Then, (maybe) $h_n$ converges to $h$ uniformly.
Noting that $f_n(x)=f_n(x)-f_n(0)=\displaystyle\int_0^x f'_n(t) dt =\int_0^x h_n(t) dt$, I get \begin{align} \left| f_n(x)-\int_0^x h(t) dt \right| &=\left| \int_0^x (h_n(t)- h(t)) dt \right| \\ &\leqq \left|\int_0^x \left|h_n(t)- h(t) \right|dt \right|. \end{align}
Since (maybe) $h_n \to h$ uniformly, I expect that $f_n(x)\to \displaystyle\int_0^x h(t) dt$ uniformly. And $g(x):=\displaystyle\int_0^x h(t) dt$ seems to work.
The problems are these.
(i) Does $h_n$ actually converge to $h$ uniformly ?
(ii) How can I firmly show $f_n(x)\to \displaystyle\int_0^x h(t) dt$ uniformly ?
$ \sum\limits_{k=1}^{\infty} a_k \sin (kx)$ converges uniformly by M-Test since $|a_k \sin kx)| \leq \frac M {k^{3}}$ and $\sum \frac M {k^{3}}<\infty$.
$ \sum\limits_{k=1}^{\infty} ka_k \cos (kx)$ converges uniformly by M-Test since $|ka_k \cos (kx)| \leq \frac M {k^{2}}$ and $\sum \frac M {k^{2}}<\infty$.
Lemma
If $f_n \to f$ uniformly and $f_n' \to g$ uniformly then $f$ is differentiable and $f'=g$.
Proof: $f_n(x)=f_n(0)+\int_0^{x} f_n'(t)dt$. Taking limit on both sides we get $f(x)-f(0)=\int_0^{x} g(t) dt$. Also, $g$ is continuous (by uniformity of convergence). Hence, $f$ is differentiable and $f'=g$.
Can you finish? [Take $ f_n(x)=\sum\limits_{k=1}^{n} a_k \sin (kx)$].