I have the following recurrence relation at hand: ( $n,m$ are integers)
$$(n+h_{n+1}) f(n+1,m) + f(n-1,m) + h_n f(n,m-1) + (m+1) f(n,m+1)=(1+m+(n-1)+ 2h_n) f(n,m) $$
where $h_{n}= \frac 1{n+1}$. We have the initial conditions, $f(n,m)=0$ if either $n<0$ or $m<0$. I will let $f(0,0)$ free.
I tried assuming detailed balance by solving the system: $$ \begin{array}{rl} (n+h_{n+1}) f(n+1,m)&= f(n,m) \\ (m+1) f(n,m+1) &= h_n f(n,m)\end{array}$$ but it seems that only $f(n,m)\equiv 0$ satisfies both of them simultaneously.
Is there another technique? Can it be solved by generating functions?
From, $$(n+h_{n+1})f_{n+1,m}=f_{n,m}-->(1)$$ $$(m+1)f_{n,m+1}=h_nf_{n,m} ---> (2)$$
We get from (2) replacing $n$ with $n+1$, $$(m+1)f_{n+1,m+1}=h_{n+1}f_{n+1,m}-->(3))$$ (1) + (3) gives, $$nf_{n+1,m} + h_{n+1}f_{n+1, m} + (m+1)f_{n+1, m+1} = f_{n,m} + h_{n+1}f_{n+1,m}$$ cancelling $h_{n+1}f_{n+1,m}$ both sides gives, $$nf_{n+1,m} + (m+1)f_{n+1, m+1} = f_{n,m} --->(4)$$ Now consider a system for $n,m >0$ from (4), $$nm!f_{n+1,m} + (m+1)!f_{n+1, m+1} = m!f_{n,m}$$ $$(n-1)(m-1)!f_{n,m-1} + m!f_{n, m} = (m-1)!f_{n-1,m-1}$$ $$(n-2)(m-2)!f_{n-1,m-2} + (m-1)!f_{n-1, m-1} = (m-2)!f_{n-2,m-2}$$ $$(n-3)(m-3)!f_{n-2,m-3} + (m-2)!f_{n-2, m-2} = (m-3)!f_{n-3,m-3}$$ $$....$$ $$1!1!f_{2,1} + 2!f_{2,2} = 1!f_{1,1}$$ $$0!0!f_{1,0} + 1!f_{1,1} = 0!f_{0,0}$$
Add all of them both sides, we end up with $$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n,m-1} + m!f_{n,m} + (n-2)(m-2)!f_{n-1, m-2} + (m-1)!f_{n-1, m-1} + .... = m!f_{n,m} + (m-1)!f_{n-1,m-1} + ... + 1!f_{1,1} + 0!f_{0,0}$$ Cancelling appropriate terms both sides, we get $$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n, m-1} + (n-2)(m-2)!f_{n-1, m-2} + ...+ 0!0!f_{1,0} = 0$$ Since the left terms must all add up to $0$ and the factorials and the product with $n,n-1..$ terms are all positive, all the $f$ terms must be $0$ resulting in $f_{n,m} = 0$ from (1)