Imagine the next identities that holds simultaneously for a given real positive $x_0$
$$A(x_0)=\sum\limits_{n=0}^{\infty}\frac{x_0^n}{n!}a_n = 0$$ $$A(-x_0)= 0$$
It is obvious to me the next identities and reasoning can follow, basically splitting the even powers to one side, and the odd powers to the other:
$$A(x_0)=\sum\limits_{n=0}^{\infty}\frac{x_0^{2n}}{(2n)!}a_{2n} + \sum\limits_{n=0}^{\infty}\frac{x_0^{2n+1}}{(2n+1)!}a_{2n+1} = 0$$ $$A(-x_0)=\sum\limits_{n=0}^{\infty}\frac{x_0^{2n}}{(2n)!}a_{2n} - \sum\limits_{n=0}^{\infty}\frac{x_0^{2n+1}}{(2n+1)!}a_{2n+1} = 0$$
So can I conclude then that necessarily: $$\sum\limits_{n=0}^{\infty}\frac{x_0^{2n}}{(2n)!}a_{2n} =0$$ and $$ \sum\limits_{n=0}^{\infty}\frac{x_0^{2n+1}}{(2n+1)!}a_{2n+1} = 0$$ ?
I totally understand from Riemann rearrangement theorem of conditionally convergent series and the way the series convergence is calculated yields to getting different results, arbitrary different results, based on the permutation put in place, basically because the partial summation is artificially manipulated.
However it is also true that if we had the capacity to expand the infinite serie, the commutative property applies indefinitely.
What are your thoughts ? Do you find a better way or reference to prove on a similar case, that odd an even series should be simultaneously $0$?