Suppose that you have a Markov chain with state space $E$ containing $0$. Assume that $$ p_{00}^{(2n)}=\binom{2n}{n}\left(\frac{1}{2}\right)^{2n-1}~~~\text{ and }~~~p_{00}^{(2n-1)}=0~~~\text{ for }n\in\mathbb{N}. $$ Is $0$ transient, positive recurrent or null recurrent?
Already showed that $0$ is recurrent.
Claim: $0$ is null recurrent.
Therefore have to show $E_0(t(0))=\infty$, where $t(0)$ is the first return time of $0$.
Question: Is it right that $$ P_0(t(0)=2k)=\frac{1}{k}\binom{2(k-1)}{k-1}\left(\frac{1}{2}\right)^{2k-1}? $$
I continued with the following:
\begin{align} \mathbb{E}_0(t(0))&=\sum_{k\geq 1}2k\cdot\underbrace{\mathbb{P}_0(t(0)=2k)}_{=\frac{1}{k}\binom{2(k-1)}{k-1}\left(\frac{1}{2}\right)^{2k-1}}\\ &=1+2\sum_{k\geq 2}\binom{2(k-1)}{k-1}\left(\frac{1}{2}\right)^{2k-1}. \end{align}
Using Stirling's formula $n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ I get \begin{equation*} 1+2\sum_{k\geq 2}\frac{1}{\sqrt{\pi (k-1)}}\cdot 2^{2(k-1)}\cdot \frac{2}{2^{2k}}=1+\sum_{k\geq 2}\frac{1}{\sqrt{\pi (k-1)}}=1+\sum_{k\geq 1}\frac{1}{\sqrt{\pi k}}\geq 1+\frac{1}{\pi}\sum_{k\geq 1}\frac{1}{k}=\infty. \end{equation*} Thus $\mathbb{E}_0(t(0))=\infty$. That is $0$ is null recurrent.
