Determine if the following Markov chain is positive recurrent, null recurrent or transcient We consider the Markov Chain with transition probabilities 
$$
p(i,0)=\frac{1}{i^2 +2},\qquad p(i,i+1)= \frac{i^2 +1}{i^2 +2}.
$$
Determine if this Markov chain is positive recurrent, null recurrent  or transcient.

My attempt: Since all states are connected to $0$, then it is sufficient to determine if $0$ is a positive recurring state.
Consider $T_{0}$ the hitting time, that is
$$T_{0}=\inf\left\{m\geq 1\: :\: X_{m}=0\right\}.$$
Note that 
$$
\mathbb{P}(T_{0}=n|X_{0}=0)=\left(\frac{1}{2}\times\frac{2}{3}\times\cdots\times\frac{(n-2)^2+1}{(n-2)^{2}+2}\right)\left(\frac{1}{(n-1)^{2}+2}\right)
$$
Therefore, we have
$$
\mathbb{E}(T_{0}|X_{0}=0)=\sum_{n=1}^{\infty}n\times \left(\frac{1}{2}\times\frac{2}{3}\times\cdots\times\frac{(n-2)^2+1}{(n-2)^{2}+2}\right)\left(\frac{1}{(n-1)^{2}+2}\right).
$$
I need to determine if this series converges or diverges. I have tried to limit it superiorly and inferiorly but I have not found good bounds.

Handle the product in the summation by taking logs:
$$\eqalign{
\log \prod_{i=0}^{n-2} \frac{i^2+1}{i^2+2} &= \sum_{i=0}^n \log\left(1 - \frac{1}{i^2+2}\right)\\
&\ge -\sum_{i=0}^n \frac{1}{i^2+2} \\
&\gt -\sum_{i=0}^\infty \frac{1}{i^2+2} \\
&\gt -\frac{1}{2} - \sum_{i=1}^\infty \frac{1}{i^2} \\
&= -\frac{3+\pi^2}{6}.
}$$
Consequently you can underestimate the sum as 
$$\eqalign {
\sum_{n=1}^\infty n \prod_{i=0}^{n-2} \frac{i^2+1}{i^2+2} \frac{1}{(n-1)^2+2} & \gt \frac{1}{e^{(3+\pi^2)/6}} \sum_{n=1}^\infty \frac{n}{(n-1)^2+2}.
}$$
The right hand side diverges (compare it to $\int_1^\infty \frac{x}{x^2+2}\mathrm{d}x$).
