Let $X_n$ be a simple, symmetric random walk on $\mathbb{Z}$ with $X_0=0$. Let $$ T=\inf\{n\ge 1 : X_n=0\} $$ Compute $\mathbb{E}(s^T)$ for fixed $s\in(0,1)$.
Apologies if this has been asked before, it seems like the sort of question that may have but not that I've found.
We have already shown that the walk is null recurrent.
I thought to use the identity $$ P_{i,j}(s)=\delta_{ij}+F_{i,j}(s)P_{j,j}(s) $$ setting $i=j=0$ and rearranging $$ \mathbb{E}(s^T)=F_{0,0}(s)=\frac{P_{0,0}(s)-1}{P_{0,0}(s)} $$ where $$ P_{0,0}(s)=\sum^{\infty}_{n=0}s^n\mathbb{P}(X_n=0)=\sum^{\infty}_{k=0}\Big(\frac{s}2\Big)^{2k}{2k\choose k} $$ and I don't see how this is useful or that it even converges.
Preferably I would like an answer that doesn't use Catalan numbers (although that may be useful for answer checking)
Thank you
First, note that $$ \mathsf{P}(T=2n)=\frac{2^{-2n}}{2n-1}\binom{2n}{n}. $$ Thus, for $s\in(0,1)$, $$ \mathsf{E}s^T=\sum_{k\ge 1}\frac{(s/2)^{2n}}{2n-1}\binom{2n}{n}=1-\sqrt{1-s^2}. $$
To see the last equality, notice that $$ A\equiv\sum_{n\ge 1}\frac{(s/2)^{2n}}{2n-1}\binom{2n}{n}=\left(\frac{s}{2}\right)^2\sum_{n\ge 1}\frac{(s/2)^{2n-2}}{2n-1}\binom{2n}{n} $$ and $$ \int_0^s \left(\frac{x}{2}\right)^{2n-1}\, dx=\frac{s(s/2)^{2n-2}}{2n-1}. $$ Consequently, since $$ \sum_{n\ge 1}\binom{2n}{n}x^{2n-2}=\frac{4}{\sqrt{1-x^2}\left(1+\sqrt{1-x^2}\right)}, $$ we get $$ A=\frac{s}{4}\int_0^s\frac{4}{\sqrt{1-x^2}\left(1+\sqrt{1-x^2}\right)}\,dx=1-\sqrt{1-s^2}. $$