We often ignore convergence when manipulating formal power series. We can add, substract, multiply, divide, differentiate, and do functional composition on such formal sums without worrying about convergence and we derive an equation formally true, well. But when we try to plug some arbitrary number into these equations, the problem happens.

Take an example from *Concrete Mathematics* EXERCISE 7.32:

> 
> An *arithmetic progression* is an infinite set of integers
>  \[\{an+b\} = \{b,a+b,2a+b,3a+b,\ldots\}\]
>  A set of arithmetic progressions $\{a\_1n+b\_1\},\ldots,\{a\_mn+b\_m\}$ is called an *exact cover* if every nonnegative integer occurs in one and only one of the progressions. For example, the three progressions $\{2n\},\{4n+1\},\{4n+3\}$ constitute an exact cover. Show that if $\{a\_1n+b\_1\},\ldots,\{a\_mn+b\_m\}$ is an exact cover such that $2 \le a\_1 \le \cdots \le a\_m$, then $a\_{m-1} = a\_m$.
> 
> 
> 

Well, we can use generating function. First assume that $b\_k \ge 0$. If $A$ is a nonneggative integer set, we consider a corresponding power series
\[
\sum\_{n \ge 0} z^n = \sum\_{k=1}^m \sum\_{n \ge 0} z^{a\_kn+b\_k} \tag{1}
\]
Formally, we derive that
\[
\frac 1 {1-z} = \sum\_{k=1}^m \frac {z^{b\_k}} {1-z^{a\_k}} \tag{2}
\]
We want to plug $\omega = e^{2\pi\imath/a\_m}$ into equation (2), but wait a minute. We find that $\sum\_{n \ge 0} \omega^n$ might diverge, and it's illegal to plug $z = \omega$ into equation (1), so we should observe something else. Well, it's trivial: eqaution (1) is true whenever $|z| < 1$, as well as equation (2). Both sides of equation (2) are rational functions which hold for all $|z| < 1$, so it holds whenever $z \in \mathbb{C}$. Now we can conclude that $a\_{m-1} = a\_m$, otherwise let $z \to \omega$, the left side of equation (2) is finity and the other is infinity.

For many readers don't know what I mean, I take another example:

Let $a\_0 = 1, a\_1, \ldots$ is a real sequence, and $A(z) = \sum\_{n \ge 0} a\_nz^n$ is its generating function. Now let's discover its reciprocal.

Assume that $A(z) = 1 + \bar A(z)$, we get

\begin{align\*}
\dfrac 1 {A(z)}
&= 1 + \sum\_{m > 0} (-1)^m \bar A(z)^m \\
&= 1 + \sum\_{m > 0} (-1)^m \sum\_{k\_1, \ldots, k\_m > 0} a\_{k\_1} \ldots a\_{k\_m} z^{\displaystyle a\_{k\_1} + \cdots + a\_{k\_m}} \\
&= 1 + \sum\_{n > 0} z^n \sum\_{m > 0} (-1)^m \sum\_{\scriptstyle k\_1, \ldots, k\_m > 0 \atop \scriptstyle k\_1 + \cdots + k\_m = n} a\_{k\_1} \ldots a\_{k\_m} = B(z)
\end{align\*}

I have interchanged $\sum$-notation without consideration of absolute convergence, so $1/A(z) = B(z)$ is only *formally* true.

Well, only from such proof, can we say that, whenever $A(z\_0)$, $B(z\_0)$ converges to a non-zero value, $1 / A(z\_0) = B(z\_0)$?

Now let's consider a more general problem. It's **what I want to ask**.

**If we get $A(z) = B(z)$ from manipulating (including adding, substracting, multiplying, dividing, differentiating, integrating and changing variables -- functional composition) generating functions (usually formal power series), and $A(z\_0)$, $B(z\_0)$ are convergent (if it is a explicit power series) or nondegenerate (if it isn't a explicit power series), can we conclude that $A(z\_0) = B(z\_0)$? If not, you could try to get some useful sufficient conditions.**

In the first example, $A(z) = 1/(1-z)$ and $B(z) = \sum\_{k=1}^m z^{b\_k}/(1-z^{a\_k})$. We get $A(z) = B(z)$ from manipulating formal power series, and it's an identity over all $z$.

Thanks for help.