Describe the behaviour of the sequence $(a_n)$ for $n\to\infty$ with $$a_n=\int_{0}^{\pi}f(x)\sin(nx)dx$$ and $f(x)$ an arbitrary continously differentiable function.
Can I argue, that with the Mean Value Theorem
$$\int_{0}^{\pi}\sin(xn)f(x)dx=f(\xi)\int_{0}^{\pi}\sin(xn)dx=f(\xi)\frac{1-cos(\pi n)}{n}$$
the sequence converges
$$\lim_{n\to\infty}(a_n)=\lim_{n\to\infty}\bigg [\ \frac{1-cos(\pi n)}{n}\bigg ]f(\xi)\to 0$$ ?