Odd triangle wave $\text{t}(x)$ with angles at $(2x+1)\in\mathbb{Z}$ can be represented by Fourier series: $$\text{t}(x)=\frac{8}{\pi^2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n-1)^2}\sin\left(\frac{(2n-1)\pi x}{2}\right)$$ And sine function can be expanded to power series the following way: $$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
As $\text{t}(x)$ is not analytic at $(2x+1)\in\mathbb{Z}$, its direct Taylor series would converge to $\text{t}(x)$ only for $x\in(-1,1)$. In fact, its Taylor series looks like: $$\text{t}(x)=x$$
This leads to an idea: what would a power series of partial Fourier series sum look like?
It is easy to see that this sum looks like this:
$$\text{t}(x)\approx\frac{8}{\pi^2}\sum_{n=1}^N\sum_{k=0}^\infty\frac{(-1)^{n+1}}{(2n-1)^2}\frac{(-1)^k}{(2k+1)!}\left(\frac{(2n-1)\pi x}{2}\right)^{2k+1}$$
It is easy to see that for finite $N$ this sum converges because $\sin(x)$ power series converges and we just do a finite sum of such series.
As $N$ increases, the power series coefficients (after we switch order of summing) for $k>0$ become larger and seem to increase indefinitely while coefficient for $k=0$ becomes closer to $1$.
I'd expect higher order coefficients to get smaller to become zero in the limit. So I wonder: how in the limit of $N\to\infty$ could they form exact $0$ so that the series becomes just $x$? Can this limit be found after some regularization? Or have I made some logical mistake?