Why higher frequencies in Fourier series are more suppressed than lower frequencies? One can expand any periodic function in sines and cosines. When calculating the coefficients $a_0$, $a_n$, and $b_n$ one find that $a_1>a_2>...>a_n>...$, similarly for $b_n$. 
Is there an intuitive reason to understand this, I mean why would one expect this to happen? It looks miraculous and mysterious to me that the coefficients came out "ordered" in a way that made the convergence manifest.
 A: 1) One may define the Fourier coefficients $a_n$ and $b_n$ (or the complex Fourier coefficients $c_n$) of a periodic function $f(\theta)=f(\theta +2\pi)$ if $f\in {\cal L}^1(\mathbb{R}/ 2\pi\mathbb{Z})$. 
2) Here ${\cal L}^1$ is an example of an ${\cal L}^p$ space, where $p\geq 1$ is a power. By definition, a function $f$ belongs to a ${\cal L}^p$ space if $f$ is measurable and the integral $\int|f|^p <\infty$ is finite. For instance, $f(\theta)=\tan(\theta)^{\frac{2}{3}}$ is an example of an ${\cal L}^1$ function that is not an ${\cal L}^2$ function.
3) It is in general not true that the Fourier coefficients $a_n$ monotonically decrease as 
$$ |a_0| \geq |a_1| \geq |a_2| \geq \ldots \geq |a_n| \geq \ldots. $$ 
It is easy to come up with counterexamples, e.g., $f(\theta)=\cos(2\theta)$ has $0=|a_1| < |a_2|\neq 0. $
4) Nevetheless, it is true that 
$$\lim_{n\to \infty}a_n = 0, \qquad \lim_{n\to \infty}b_n = 0, \qquad \lim_{|n|\to \infty}c_n = 0,$$
if $f\in {\cal L}^1(\mathbb{R}/ 2\pi\mathbb{Z})$. This is a consequence of Riemann–Lebesgue lemma.
A: The reason of a more modest version of your statement (your big claim is not right) is that the sum
$$\sum_{n=-\infty}^{\infty} |a_n|^2 $$
has to converge. That's because this sum is proportional to 
$$ \int_0^{2\pi} |f(x)|^2 dx $$
which converges for bounded functions (a basic insight about Fourier expansions and Hilbert spaces of periodic functions). That's why, for example, power law
$$ |a_n| \sim \frac{1}{n^\epsilon} $$
require $\epsilon>0.5$ or, if some logarithmic corrections are included, at least $\epsilon\geq 0.5$. If the Fourier coefficients were not dropping at least this quickly, the function wouldn't be $L^2$-integrable: the sum above wouldn't converge.
Of course, this convergence requirement doesn't prevent some coefficients from being larger than $C/n^\epsilon$ as long as most others drop quickly. Still, there can't be infinitely many coefficients $a_n$ such that $a_n > \delta$ for a pre-given positive $\delta$ because the sum would still diverge.
Distributions which are not really functions such as $f(x)=\delta(x)$ may have Fourier coefficients that don't drop: of course, $\delta(x)^2$ doesn't have a finite integral, anyway. The same holds for unbounded functions.
The condition $a_1 > a_2 > a_3 > \dots$ is much stronger and is obviously violated for "most functions": you may simply choose coefficients that don't agree with this strict ordering and construct a corresponding function. Still, simple enough functions tend to obey even this strict ordering because the calculation of $a_n$ leads to a simple enough and monotonic function of $n$. 
Perfectly smooth functions have $a_n$ decreasing faster than any power law; functions with steps have $a_n\sim 1/n$; continuous functions with $|x|$-like unsmooth points have $a_n\sim 1/n^2$ for large $n$, and so on.
