Consider a chain with state space $\{1,2, \cdots \}.$ If you are at $1$ go to state $j$ with probability $p_j$ $($$ j=1,2,\cdots$ $) ,$ where these are non-negative numbers adding to $1$. If you are in a state $i>1\ ,$ then go just one step back$,$ that is$,$ to $i-1.$ Discuss the nature of the states and nature of the stationary distribution$.$
Assuming that all $p_j$'s are positive I found that the Markov chain is irreducible with all it's states recurrent$.$ So there is no transient state and hence inessential state in the above Markov chain$.$ While calculating stationary distribution I found a problem$.$ Here it is $:$
Suppose $\pi = (\pi_1 , \pi_2 , \cdots )$ be the stationary distribution of the above Markov chain$.$ Then I found that $\pi_2 = (1-p_1) \pi_1, \pi_3 = (1-p_1-p_2) \pi_1, \cdots$ Since $\sum\limits_{i=1}^{\infty} \pi_i = 1$ so $\pi_1 \{1 + (1-p_1) + (1-p_1 -p_2) + (1-p_1 -p_2 - p_3) + \cdots \}= 1.$ Now how do I find the sum
$$1 + \sum_{k=1}^{\infty} (1-p_1-p_2 -p_3 - \cdots - p_k)\ ?$$
Or in other words
$$\sum\limits_{k=1}^{\infty} \sum\limits_{j=k}^{\infty} p_j\ ?$$
This gives us $\pi_1$ and consequently all the $\pi_i$'s$.$ where
$$\pi_i = \frac {\sum\limits_{k=i}^{\infty} p_k} {\sum\limits_{k=1}^{\infty} \sum\limits_{j=k}^{\infty} p_j}.$$
for $i=1,2,\cdots.$
Please help me in this regard. Thank you very much.
$$\begin{align} \sum_{j=1}^\infty{jp_j}&=\sum_{j=1}^\infty{p_j}+\sum_{j=2}^\infty{(j-1)p_j}\\ &=1+\sum_{j=2}^\infty{p_j}+\sum_{j=3}^\infty{(j-2)p_j}\\ &=1+(1-p_1)+\sum_{j=3}^\infty{p_j}+\sum_{j=4}^\infty{(j-3)p_j}\\ &\vdots \end{align}$$
EDIT
This is true if the given sum converges. As discussed in the accepted answer to this question, it need not.