A zero-seeking device operates as follows: if it is in state $j$ at time $n$, then at time $n+1$, its position is $0$ with probability $\frac{1}{j}$ or $k$ with probability $\frac{2k}{j^2}$, where $k$ is one of the states $1,2,...,j-1$. State $0$ is defined to be absorbing.
Now, I am defining $W_{i,j} = P(\text{Ever visiting state} \space $j$ \space \space|\space \space X_0 =i)$. Then, clearly $W_{i,i} = 1$ and $W_{i,j}=0$ for $j>i$. The formula for $W_{k,j}$ for $k>j$ is given without proof in my book to be:
$W_{k,j} = 2\frac{k+1}{k}\frac{j}{(j+1)(j+2)}$ for $1\leq j <k$.
However, I am not sure how this result was derived. Would anyone be able to give me any pointers? Thank you!
Let $p_{jk}=2k/j^2$ be the one step transition probability.
Then from state $m+1$ we can only get to state $m$ if we jump there in the next step, so $W_{m+1,m}=p_{m+1,m}=2m/(m+1)^2.$
Note that this expression satisfies the general formula for $W$ so we can start the induction here and continue to $W_{m+2,m}, W_{m+3,m},...$
Starting from state $j$ we condition on the first step. $$ W_{j,m}=\sum_{k=m+1}^{j-1}p_{j,k}W_{k,m}+p_{j,m}W_{m,m} $$ We have to be careful here since we want to take $W_{m,m}=1$ but the general formula does not yield $1.$ Now use the induction hypothesis on the $W_{k,m}$ in the sum and compute the sum. (Most of the terms in the sum are constants.)