The proof of the eventual return to zero of a symmetric random walk was given here, but I am not comfortable with generating functions.
In this book chapter an intermediate result depends on generating functions: In deriving the probability of a first pass through origin at a time $2m$ as:
$$f_{2m}=\frac{u_{2m}}{2m-1}=\frac{{2m\choose m}}{(2m-1)2^{2m}}$$
where $u_{2m}$ is the probability of return to origin at time $2m$: $u_{2m}={2m\choose m}2^{-2m}$, the following generating functions are defined:
$$U(x)=\displaystyle \sum_{m=0}^\infty u_{2m}\,x^m$$ $$F(x) = \displaystyle \sum_{m=0}^\infty f_{2m}\, x^m$$
with the following relation: $U(x) = 1 +U(x)F(x)$.
A closed form expression of $U(x)$ is found: $U(x)=\frac{1}{\sqrt{(1-x)}}$, arriving at $F(x)=1-(1-x)^{1/2}$.
I don't understand the following step, which is explained as:
"Although it is possible to compute the value of $f_{2m}$ using the binomial theorem, it is easier to note that $F'(x)=U(x)/2$, so that the coefficients of $f_{2m}$ can be found by integrating the series for $U(x)$. We obtain, for $m\geq1$,
$$f_{2m}=\frac{u_{2m}-2}{2m}=\frac{{2m-2\choose m-1}}{m2^{2m-1}}=\frac{{2m\choose m}}{(2m-1)2^{2m}}=\frac{u_{2m}}{2m-1}."$$
Can I get help breaking down the steps right after the sentence in bold? An actual step-by-step derivation would be very good, but a recommendation for an "accessible" book or article dealing with random walks, or other suggestions as to how to understand the use of generating functions in this context would be equally welcome.
After looking up generating functions, it seems like the answer would go as follows:
The derivative of $F(x)=1-(1-x)^{1/2}$ is $\large F'(x)=1/2(1-x)^{-1/2}= \frac{1}{2\sqrt{1-x}}$. Since $\large U(x) = \frac{1}{\sqrt{1-x}}$, it follows that $F'(x) =\frac{U(x)}{2}$. If we want the coefficients for $F(x)$ we can integrate:
$$F(x) = 1/2\Big(\frac{u_o x^1}{1}+\frac{u_2 x^2}{2}+\frac{u_4 x^3}{3}+\frac{u_6 x^4}{4}+\cdots\Big)=\frac{1}{2}\displaystyle\sum_{m=1}^\infty \frac{u_{2m-2}\,\,x^m}{m}$$Therefore the equivalence between coefficients will be:
$$f_{2m}=\frac{u_{2m-2}}{2m}.$$