Let $f(x)$ be a smooth function. What is the geometric intuition for why $\int_{0}^{2\pi}f(x)\sin(nx) < \int_{0}^{2\pi}f(x)\sin(mx)$ for $nLet $f(x)$ be a smooth function. What is the geometric intuition for why $$\int_{-\pi}^{\pi}f(x)\sin(nx) < \int_{-\pi}^{\pi}f(x)\sin(mx)$$ for $n<m$?
If $f(x)$ is a smooth function, then $f(x)$ equals it's fourier series:
$$f(x) = \frac{a_0}{2}+\sum_{k=1}a_n\cos(kx)+b_n\sin(kx)dx$$
Where
$$a_k = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(kx)dx$$
$$b_k = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(kx)dx$$
for $k \geq 1$ are the Fourier coefficients of $f$. I believe that the convergence of this sequence implies that the coefficients $a_k$ and $b_k$ are eventually decreasing;
$\int_{-\pi}^{\pi}f(x)\sin(nx) < \int_{-\pi}^{\pi}f(x)\sin(mx)$ for $n<m$ (and similarly for $\cos(kx)$).
But then again, the identity $\sum_{k=1}^{N}\cos(kx) = \dfrac{\sin(Nx+\frac{x}{2})}{2\sin(\frac{x}{2})}-\frac{1}{2}$  (and a similar identity for $\sum_{k=1}^{N}\sin(kx))$ make me not completely sure that the coefficients of the Fourier series are eventually decreasing...
Sorry, my question sort of evolved as I wrote this $>.<$... If anyone can spread some insight into my original question (or anything else I brought up here) I'd appreciate it.

Edited due to jrbr's response:
is there is a geometric intuition for why $\int_{-\pi}^{\pi} f(x)cos(nx) \rightarrow 0$ as $n \rightarrow \infty$
 A: Pick $\epsilon>0$. Since $f$ is uniformly continuous, I can pick a $\delta>0$ such that $|x-y| < \delta$ implies $|f(x) - f(y)| < \frac \epsilon 2$. Now suppose that $f$ is positive. Pick $n > \frac \pi \delta$. Then
\begin{align}
\int_0^{2 \pi} f(x) \sin( nx) dx &= \sum_{k=0}^{n-1}\int_{\frac{2k}{n}\pi}^{\frac{2k+1}{n}\pi} f(x) \sin( nx) dx + \int_{\frac{2k+1}{n}\pi}^{\frac{2k+2}{n}\pi} f(x) \sin( nx) dx\\
& \le  \sum_{k=0}^{n-1}\int_{\frac{2k}{n}\pi}^{\frac{2k+1}{n}\pi}\left( f\left (\frac{2k+1}{n}\pi\right ) + \epsilon\right)\sin( nx) dx + \int_{\frac{2k+1}{n}\pi}^{\frac{2k+2}{n}\pi} \left( f\left (\frac{2k+1}{n}\pi\right )- \epsilon \right) \sin( nx) dx\\
&=  \sum_{k=0}^{n-1}\frac 1n \left( f\left (\frac{2k+1}{n}\pi\right ) + \epsilon\right)-\frac 1n \left( f\left (\frac{2k+1}{n}\pi\right )- \epsilon \right) \\
&=\epsilon.
\end{align}
A very similar calculation can also be done to show that
$$
\int_0^{2 \pi} f(x) \sin( nx) dx \ge - \epsilon.
$$
This proves the conclusion you made before about limits (for positive $f$). The geometric intuition that it represents is that, when the frequency of the sine wave gets high, $f$ doesn't vary very quickly in comparison, and so the positive and negative bits of each sine wave mostly cancel out in the integral. What I presented can be extended to $f$ with both positive and negative parts, but the calculation just gets a bit uglier.
