Suppose $f$ is differentiable any number of times and is $2\pi$-periodical. Let $k\in \mathbb{N}$. Show that $$\lim_{n\to\infty}n^k|f(x)-s_n(x)|= 0$$ i.e that $f(x)=s_n(x)+o(1/n^k)$ as $n\to \infty$. The $s_n$ are the $n$th Fourier partial sums.
Intuitively I would like to show this using an argument along the following lines: \begin{align*} s_n(x) - f(x) &= \int_{-\pi}^{\pi}[f(x+y)-f(x)]D_n(y)dy\\ &= \int_{\delta\leq |y| \leq \pi}[f(x+y)-f(x)]D_n(y)dy+\int_{0\leq |y|\leq \delta}[f(x+y)-f(x)]D_n(y)dy\\ &\approx 0 + \int_{0\leq |y|\leq \delta}o\Big(\dfrac{1}{n^k} \Big)D_n(y)dy \end{align*} as $n\to\infty$, for some $\delta>0$ small enough (since $f$ is differentiable). However, this seems hard to make rigorous, and having both sides depend on $n$ makes me hesitant to use this little-o notation. Some hints/solutions would be appreciated.
$f=\sum_n c_n(f) e^{inx}$ and $(in)^{k+2} c_n(f)= c_n(f^{(k+2)})$ is bounded by $\|f^{(k+2)}\|_{L^1[0,2\pi]}$ thus $$|\sum_{|n|> N} c_n(f)e^{inx}|\le N^{-k-1}\|f^{(k+2)}\|_{L^1[0,2\pi]}$$