Fourier sine series: $$ f(x) = \sum_{n=1}^{∞}s_n(f)\sin(nπx), \;\:\, s_n(f) = 2 \int_{0}^{1}\sin(nπx)f(x)dx,$$ Fourier cosine series: $$ f(x) = \frac{c_0}{2} + \sum_{n=1}^{∞}c_n(f)\cos(nπx), \;\:\, c_n(f) = 2\int_{0}^{1}\cos(nπx)f(x)dx. $$ For given $k ∈ \mathbb{N_0}$, show that $$\sum_{n=1}^{∞} n^k|c_n(f)| < ∞, \;\:\, \sum_{n=1}^{∞}n^k|s_n(f)| < ∞ $$ if $f ∈ C^{k+1}([0, 1], \mathbb{C})$ with $f^{(2j)}(0) = f^{(2j)}(1) = 0$ for $0 ≤ j ≤ k/2$ for the case of $s_n(f)$ and $f^{(2j+1)}(0) = f^{(2j+1)}(1) = 0$ for $0 ≤ j < k/2$ for the case of $c_n(f)$.
My Approach:
$$ c_n(f') = 2\int_{0}^{1} \cos(n\pi x)f'(x)dx = 2[\cos(nx\pi)f(x)]_{0}^{1} - 2\int_{0}^{1} \cos'(n\pi x)f(x)dx = 2[\cos(n\pi)f(1)-f(0)] + n\pi\int_{0}^{1} \sin(nx\pi)f(x)dx = 2((-1)^nf(1)-f(0)) + n\pi s_n(f). $$
Furthermore
$$ s_n(f') = 2\int_{0}^{1} \sin(n\pi x)f'(x)dx = 2[\sin(nx\pi)f(x)]_{0}^{1} - 2\int_{0}^{1} \sin'(n\pi x)f(x)dx = 2[-n\pi\int_{0}^{1}cos(n\pi x)f(x)dx = -n\pi c_n(f) $$
My Question:
Am I on the right way to a solution? And is it possible that that should be it?