In Sec. 10.12 (hitting times of simple random works), eq. (c) (page 103) of Probability with Martingales (D. Williams 1991), the author said:
For (any) $\theta\in \mathbb{R}$, let $\alpha:=\mathrm{sech}(\theta)$, we have: $$\sum_{n=1}^{\infty} \alpha^n \mathbb{P}(T=n)=\alpha^{-1}[1-\sqrt{1-\alpha^2}]$$ so that $$\mathbb{P}(T=2m-1)=(-1)^{m+1}\binom{\frac{1}{2}}{m}$$ $T$ is the stopping time of the simple random work, i.e., a discrete random variable taking $\{1,3,5,...\}$
How did he get such a result?
My conjecture: one may let (say) $\alpha=0,0.1,0.01,...,1 \in[0,1]$ to create infinitely many equations and solve each (of the infinite many) variable $\mathbb{P}(T=2m-1)$ for $m=1,2,3,...$ in these equations.
Besides, I do not really understand the notation $\binom{\frac{1}{2}}{n}$. If it denotes the combination number, how can one write a fraction $\frac{1}{2}$?
For a real number $b$ and a non-negative integer $n$, the meaning of $b\choose n$ is $$ {b(b-1)\cdots(b-n+1)\over n!},\qquad n\ge 1, $$ with ${b\choose 0}=1$ being agreed upon by convention. With this notation, you have Newton's formula $$ (1+t)^b=\sum_{n=0}^\infty {b\choose n} t^n,\qquad |t|<1. $$ You can use this with $b=1/2$ and $t=-\alpha^2$ to expand $\alpha^{-1}\left[1-\sqrt{1-\alpha^2}\right]$ in a power series. Then equate coefficients of like powers of $\alpha$ to see that $$ P(T=2m-1) = (-1)^{m+1}{1/2\choose m},\qquad m=1,2,\ldots. $$