Let $(X_n), n \geq 0$ be a transient Markov chain on the non-negative integers with transition probability matrix $P = \|P_{ij}\|$. Define $u(i) = \sum_{n=0}^{+\infty} P_{i0}^{(n)}$.
Then $u(X_k)$ is submartingale.
My attempt:
Let $\Gamma = [X_1 = i_1, \dots, X_k = i_k]$.
We just need to show that $\mathbb{E}[u(X_{k+1})\textbf{1}_\Gamma] = \mathbb{E}[u(X_{k})\textbf{1}_\Gamma]$
\begin{align} \mathbb{E}[u(X_{k+1})\textbf{1}_\Gamma] &= \sum_{j=0}^{+\infty} \mathbb{E}\left[\sum_{n=0}^{+\infty} P_{j0}^{(n)} \textbf{1}_\Gamma \textbf{1}_{X_{k+1}=j}\right] \\ &= \sum_{j=0}^{+\infty} \mathbb{E} \left[\sum_{n=0}^{+\infty} \sum_{i=0}^{+\infty} P_{ji}P_{i0}^{(n-1)}\textbf{1}_\Gamma \textbf{1}_{X_{k+1}=j}\right] \\ &= \sum_{j=0}^{+\infty} \mathbb{E} \left[\sum_{i=0}^{+\infty} P_{ji} \sum_{n=0}^{+\infty} P_{i0}^{(n-1)} \textbf{1}_\Gamma \textbf{1}_{X_{k+1}=j}\right] \end{align}
As we have a Markov chain, for a fixed $j$, $\sum_{i=0}^{+\infty} P_{ji} = 1$.
Thus:
$$ \mathbb{E}[u(X_{k+1})\textbf{1}_{\Gamma}] = \sum_{j=0}^{+\infty} \mathbb{E}\left[\sum_{n=0}^{+\infty} P_{i0}^{(n-1)}\textbf{1}_\Gamma \textbf{1}_{X_{k+1}=j}\right] $$
How can I continue from here? I made some more steps,but it not lead me to anything...