A markov chain on states 0,1,... has transition probabilities $P_{ij}=\frac{1}{i+2}$ for $j=0,1,...,i,i+1$. Find the stationary distribution.
My attempt:
First I calculate the equations
$\Pi_0$=$\Pi_0/2+\Pi_1/3+\Pi_2/4+\Pi_3/5+...$ and so on
But then I don't know how to continue, I mean how to solve for $\Pi_k$.
Can somebody help me please?
Well, instead of saying "and so on" write them down. You will see that they are all very similar to one another. For instance we have $$ \Pi_0 = \frac{\Pi_0}{2}+\frac{\Pi_1}{3}+ \frac{\Pi_2}{4}+\ldots\\\Pi_1 = \frac{\Pi_0}{2}+\frac{\Pi_1}{3}+ \frac{\Pi_2}{4}+\ldots\\\Pi_2 = \frac{\Pi_1}{3}+\frac{\Pi_2}{4}+ \frac{\Pi_3}{5}+\ldots$$
So we have $\Pi_0 = \Pi_1$ and $\Pi_1 = \frac{\Pi_0}{2}+\Pi_2.$ So immediately we have $\Pi_2= \Pi_0/2.$ Now we can write $\Pi_2 = \Pi_1/3+\Pi_3$ so that $\Pi_3 = \Pi_0/6.$ Then $\Pi_3 = \Pi_2/4 + \Pi_4$ so that $\Pi_4 = \Pi_0/24.$ See the pattern?
To see the factorial come out more clearly, try adding the equations together. If you add the equations from $\Pi_{i+1}$ on you get $$ \Pi_{i+1}+\Pi_{i+2}+\ldots = \frac{\Pi_i}{i+1} + 2\frac{\Pi_{i+1}}{i+2} +3\frac{\Pi_{i+2}}{i+3}+\ldots$$ which rearranges to $$ \frac{\Pi_{i}}{i+1} = i\left(\frac{\Pi_{i+1}}{i+2} +\frac{\Pi_{i+2}}{i+3}+\ldots\right)$$ and can be further rewritten to $$ \Pi_i = i\left(\frac{\Pi_i}{i+1}+\frac{\Pi_{i+1}}{i+2} +\frac{\Pi_{i+2}}{i+3}+\ldots\right)$$ But we can plug in $i-1$ for $i$ in the first equation to give $$ \frac{\Pi_{i-1}}{i} = (i-1)\left(\frac{\Pi_i}{i+1}+\frac{\Pi_{i+1}}{i+2} +\frac{\Pi_{i+2}}{i+3}+\ldots\right) = (i-1)\frac{\Pi_i}{i}$$ where we plugged in the second equation. This simplifies to $$ \Pi_i = \frac{\Pi_{i-1}}{i-1}.$$
(There might be an easier way to get this 1st order recursion equation that I'm missing...)