Let $\alpha,\beta\in \mathbb{R}$, $E=\lbrace 0,1,2,...\rbrace$ and $Q=(q(x,y):x,y\in E)$ given by \begin{align} q(0,n)=\begin{cases} -1 &\mbox{ if } n=0 \\ 1 &\mbox{ if } n=1 \\ 0 &\mbox{ otherwise } \end{cases} \end{align} and for $m>0,$ \begin{align} q(m,n)=\begin{cases} -m^{\alpha}-m^{\beta} &\mbox{ if } n=m \\ m^{\alpha} &\mbox{ if } n=0 \\ m^{\beta} &\mbox{ if } n=m+1 \\ 0 &\mbox{ otherwise }\end{cases} \end{align} The question is to determine all $(\alpha,\beta)$ such that $Q$ is the $Q$-matrix of a non-explosive continuous time Markov chain.
First of all, the $Q$-matrix belongs to the chain with transitions $p(0,1)=1$ and, for $m>0$, $$p(m,m+1)=\frac{m^{\beta}}{m^{\alpha}+m^{\beta}},\\ p(m,0)=\frac{m^{\alpha}}{m^{\alpha}+m^{\beta}}$$ and the rates $c$ of departing from a state are $c(0)=1$ and $c(x)=m^{\alpha}+m^{\beta}$.
Since $c(x)>0$ $\forall x\in E$, then there are no absorbing states.
Next is to determine the $(\alpha,\beta)$ such that the chain is non-explosive, which can be calculated e.g. by showing $\sum_{i=1}^{\infty} 1/c(Z_i)=\infty$, where $Z_i$ is where you are at time $i$. We have \begin{align} \sum_{m=1}^{\infty} \frac{1}{c(m)} = \sum_{m=1}^{\infty} \frac{1}{m^{\alpha}+m^{\beta}}=\begin{cases} <\infty &\mbox{ for } \alpha\vee\beta>1 \\ =\infty &\mbox{ for } \alpha\vee\beta\leq 1 \end{cases} \end{align} So for $\alpha\vee\beta>1$ it completely depends on if the chain is recurrent, i.e. $$\prod_{i=1}^{\infty} \frac{m^{\beta}}{m^{\alpha}+m^{\beta}}<1$$ Is it doable to calculate this product? I tried making a sum of it by taking logarithms, and then wolframalpha says Comparison test. I don't see how that test can be applied here. Now I have an answer, but only with the help of wolframalpha, not with mathematical tools.
Is there anyone who has a hint/solution? Thanks in advance.
Regarding whether the product converges to $0$ or not, we can use the following from the Wikipedia page on convergence criteria for infinite products (which has the proof there):
If $a_m = 1 - q_m$ for some $0 \leq q_m < 1$, then $\prod_{m=1}^\infty a_m$ converges to a non-zero number iff $\sum_{m=1}^\infty q_m$ converges.
In our case we let $$a_m = \frac{m^\beta}{m^\alpha + m^\beta}, \textrm{ so set } q_m = \frac{m^\alpha}{m^\alpha + m^\beta}.$$
Then each $q_m$ satisfies $0 \leq q_m < 1$. And $$\sum_{m=1}^\infty q_m = \sum_{m=1}^\infty \frac{m^\alpha}{m^\alpha + m^\beta} = \sum_{m=1}^\infty \frac{1}{1 + m^{\beta - \alpha}} < \sum_{m=1}^\infty \frac{1}{m^{\beta - \alpha}}.$$ The sum on the right-hand side converges whenever $1 < \beta - \alpha$, so by the comparison test, $\sum_{m=1}^{\infty} q_m$ also converges when $1 < \beta - \alpha$. You can also see that $\sum_{m=1}^{\infty} q_m$ will diverge if $\beta - \alpha \leq 1$.
So the product $$\prod_{m=1}^\infty \frac{m^\beta}{m^\alpha + m^\beta}$$ converges to $0$ iff $\beta - \alpha \leq 1$.