pi approximation with Newton's method to an arbitrary rate of convergence

So, what you are doing is a Hermite interpolation, with trigonometric polynomials, of the (inverted) sawtooth wave $$ g(x) = \pi - x\quad \left| {\;x \in \left( {0,2\pi } \right)} \right. $$ $$ g(x) = 2\pi \left( {\left\lfloor {{1 \over 2} + {x \over {2\pi }}} \right\rfloor - {x \over {2\pi }}} \right) $$

You are doing a Hermite interpolation, since you consider that the function is odd, and therefore you use only sines.
After that you obtain implicitly that $f(\pi )=0$, as all its even derivatives, and impose 1st derivative to be $-1$ at $x= \pi $, and then the 3rd and 5th to be null.

Intuitively speaking, you are developing a local approximation to $g(x) \; |\, x= \pi $, a sort of Taylor series but with trigonometric polynomials in place of the normal ones.

For sure the $f_{n} (x)$ will improve the convergence of Newton method, and the improvement will be of the order of the Taylor series, since for instance $$ \sin \left( {3x} \right) = 3\cos ^2 x\sin x - \sin ^3 x = 3\sin x - 4\sin ^3 x $$ which means that you can convert your approximation into $$ f_n (x) = \sum\limits_{k = 1}^n {a_{\,k} \sin \left( {kx} \right)} = \sum\limits_{k = 1}^n {b_{\,k} \sin ^{\,k} x} = \sum\limits_{k = 1}^n {c_{\,k} x^{\,k} } + O\left( {x^{\,n + 1} } \right) $$ i.e. the difference with the truncated Taylor series will be of the order of $O(x^{n+1})$.

The Fourier series is different, as it can be thought as a global (on the whole period) approximation, in the least squares acception.

Clearly both will converge to $g(x)$ for large $n$, even because the sines ensures that your $f(x)$ is null at $x=0, 2\pi $, same as the Fourier series.

Interesting question! Here are some general comments.

1. On the one hand, your result is not really unexpected (e.g., cf. uniqueness of Fourier series). Say you want the best trigonometric approximation of $\pi-x$: $$ \pi-x=\sum_{i=0}^ka_i\sin(ix)+R_N(x) $$ A possible approximation can be obtained by insisting that both sides have the same value, and derivatives, at (say) $x=\pi$. This leads to your conditions on the coefficients. To make this precise, one must prove that $R_k(x)\to0$. But, intuitively speaking, this is indeed what you expect. Furthermore, in this limit, the coefficients should become the Fourier coefficients of $\pi-x$, to wit, $$ \lim_{k\to\infty}a_i=\frac{1}{\pi}\int_0^{2\pi}(x-\pi)\sin(ix)\mathrm dx=\frac{2}{i} $$

2. With a little bit of work, you can prove that the coefficients are actually given by $$ a_i=\frac{2 (k!)^2}{i (i+k)! (k-i)!}=\frac{2}{i}-\frac{2i}{k}+\mathcal O(k^{-2}) $$ as expected.

3. More generally, say you are interested in a series of the form $$ \sum_{i=0}^ka_i f_i(x) $$ where $\{f_i\}$ are a basis (of some reasonable function space), and the $\{a_i\}$ are fixed by some differential conditions. In the limit $k\to \infty$, if this converges to a function $f(x)$, then this function should satisfy the same differential conditions than its truncation (under some typical regularity assumptions). Here you want $f(\pi)=0$, $f'(\pi)=-1$, and all the rest of derivatives to vanish. The only (analytic) function that satisfies these conditions is indeed $f(x)=\pi-x$, as follows from uniqueness of Taylor expansions.