I've read this in a post here (can't remember which - might even have been a comment) that I thought that was the most ridiculous thing I have ever heard.
Can someone illustrate mathematically that this operation is indeed possible or impossible?
Note:
Actually when I think about it, the comment was about how as amplitudes of all sinusoids converged to zero...something around that ball park
A "zero amplitude sinusoid" isn't a sinusoid at all. It is a straight line, and the function is equal to zero everywhere, i.e. $\forall x: f(x) = 0$. This means that if $f_n$ is one such function, then
$$\sum_{n=1}^\infty f_n(x) = \sum_{n=1}^\infty 0 = 0,$$ meaning that the only function that the sum of infinitely many zero amplitude sinusoids can converge to is the constant zero function.
Even if the amplitudes of all sinusoids converges to zero, the answer will be no. We would have $f_n(x) = a_n\sin x$ where $a_n\to0$ as $n\to\infty$, but that means that for $xk=\pi k$ for any natural number $k\in\mathbb N$, you would still have
$$\sum_{n=1}^\infty f_n(x_k) = \sum_{n=1}^\infty 0 = 0$$
and indeed, you can show that any function that is the sum of $a_n \sin x$ is actually equal to $$\sin x \cdot\left(\sum_{n=1}^\infty a_n\right)$$ and so a sinusoid itself.
If you really want to approximate functions using trigonometric functions, you will have to start changing not only the amplitudes, but the frequencies of the functions. For example, you can approximate (the convergence is uniform) any function on $[0,1]$ using only functions $$\sin\left(\frac{2\pi x}{k}\right), \cos\left(\frac{2\pi x}{k}\right), k\in\mathbb N$$