I am reading "Introduction to Combinatorial Mathematics" by C. L. Liu.
The author wrote about generating functions as follow:
However, because $x$ is just a formal variable, there is no need to question whether the series converges.
The author also wrote as follows:
I think the author used the fact $\sum_{i=0}^{\infty}\binom {2i}{i}x^i$ converges for some $x\neq 0$ and $(1-4x)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\binom {2i}{i}x^i$ holds.
I think the author also used this fact.
So, I don't think "there is no need to question whether the series converges".
Am I wrong?
I think if $\sum_{i=0}^{\infty}\binom {2i}{i}x^i$ didn't converge for any $x\neq 0$, then we could not derive any useful results from this power series.
Is the power series like $\sum_{n=0}^{\infty} n!x^n$ usuful in combinatorics?

The answer to the question
is "yes". A simple example is that if you want to count connected graphs on $n$ labeled vertices, you do it using the exponential formula: if $c_n$ is the number of connected graphs, then $$ \exp\left(\sum_{n \geq 1} c_n \frac{x^n}{n!}\right) = \sum_{n \geq 0} 2^{\binom{n}{2}} \frac{x^n}{n!} $$ (because the RHS is the exponential GF for all graphs by number of vertices) and we can extract $c_n$ by taking logarithms $$ c_n = n! \cdot [x^n] \log\left(\sum_{n \geq 0} 2^{\binom{n}{2}} \frac{x^n}{n!}\right).$$ Relationships of this kind are extremely useful, even where (as here) the series diverges badly if you try to evaluate. An example involving the specific series you asked about is the enumeration of indecomposable permutations.
Here are a couple other places where some version of this question has been asked: