Can I directly substitute a specific complex number into a formal power series equation? Suppose there is a formal power series equation, such as
$$\sum_{i=0}^\infty a_i x^i=\sum_{i=0}^\infty b_i\left(\sum_{j=1}^\infty c_j x^j\right)^i.$$
If there is a complex number $r$ which makes $\sum c_jr^j$ and $\sum a_i r^i$ convergent, also, the convergent point of $\sum c_jr^j$, $s$, makes $\sum b_i s^i$ convergent, then, does the equation above still hold when I substitute $r$ into $x$?
Thanks for your discussion.
 A: The operation involves re-arranging the power series.  Hence this can only be done if the respective sums converge absolutely.  This is satisfied if the power series is evaluated at a value from inside the respective radius of convergence.
If a power series just converges conditionally, then re-arranging might change its value.  This can only happen if the value where the series is evaluated is on the boundary of the radius of convergence.
See also Riemann's rearrangement theorem.
There are also some posts on mathoverflow that deal with value-preserving permutations of conditionally convergent series, like The stabilizer of the conditionally convergent series.

ad comment: Just that a function is analytic at some point doesn't tell you anything about it's power series expansion.  Take $u(z) = \sum_0 z^n$ for example, with radius of convergence $r_u=1$. $u$ is analytic everywhere except $z=1$, but the series does not converge for any $|z|>1$.  And it does not converge for, say, $z=-1$ or $z=\sqrt{-1}$, either.
Also, if a power series expansion of $f$ around 0 converges for $z$ with $|z|=r_f$, then $f$ is countinuous at $z$.  Again, the opposite is not true in general, and continuiuty of $f$ doesn't tell you anything about $r_f$ or about the becaviour of a power series expansion.
However, your requirement that $f$ is analytic in$^1$ $0\leqslant |z|\leqslant r_f$ for $f\in\{F,G,H\}$ means that the radius of convergence of $f$ is greater than $r_f$ and $r_f$ does not denote the radius of convergence of $f$.  The reason is that, when a power series has a finite radius of convergence $r_f\neq0$, then $f$ has a singularity of some sort for at least one $|z|=r_f$.  Example is polylog $\operatorname{Li}_2(z) = \sum_1 z^n/n^2$ that converges for all $|z|=1$ but it has still a singularity: a branch point at $z=1$.
The conclusion is that you are inside$^1$ the radius of convergence of the functions, thus their power series converge absolutely.
Notice that $z$ above denotes a complex number, and if you replace it by a real number, then what's said above is no more true.  For example, $\arctan z =\sum_0 z^{2n+1}/(2n+1)$ is analytic over $\Bbb R$, yet the power series has finite radius of convergence $r_{\arctan} = 1$.  The reason is that $\arctan$ has branch points at $z=\pm\sqrt{-1}$, so you only see whan't going on over $\Bbb C$.

$^1$Assuming with $D(z,r)$ you denote the closed disk of radius $r$ around $z$.
If, however, $D$ denotes the open disk and $r_f$ stands actually for the respective radius of convergence, then analyticity of $f$ in the open disk $0\leqslant |z| < r_f$ does not tell about whether the power series converges (absolutely) for some $|z|=r_f$.
