Let $(X_n)$ be a Markov chain on a countable space $E$ and let $T$ be it's transition matrix. We assume $T(x,x)<1$ for all $x\in E$. Let $\tau=\inf\{n\geq 1:X_n\neq X_0\}$. I want to show that $\tau$ is finite $\Bbb{P}_x$ almost surely.
So my idea was to show that $\Bbb{P}_x(\tau<\infty)=1$. I thought it would be equivalent to show $\Bbb{P}_x(\tau=\infty)=0$ hence $$\begin{align} \Bbb{P}_x(\tau=\infty)&= \Bbb{P}_x(\forall x~X_n=x)\\&= \Bbb{P}_x\left(\bigcup_{n\geq 1} X_n=x\right)\\&\leq \sum_{n\geq 1}\Bbb{P}_x(X_n=x)\\&=\sum_{n\geq 1}\Bbb{P}(X_n=x|X_{n-1}=x)\\&=\sum_{n\geq 1} T(x,x)\end{align}$$
But somehow this does not give me zero. Could maybe someone help me how to show this?
We have for $k\in\mathbb N^*$: \begin{eqnarray*} \mathbb P_x(\tau=k)&=&\mathbb P_x(X_0=x, X_1=x,...,X_k=x,X_{k+1}\neq x)\\ &=&\mathbb P_x\left(\bigcup_{e\in E\backslash\{x\}}X_0=x,...,X_k=x,X_{k+1}=e\right)\\ &=&\sum_{e\in E\backslash\{x\}}\mathbb P_x(X_0=x,...,X_k=x,X_{k+1}=e)\\ &=&\sum_{e\in E\backslash\{x\}}\mathbb P_x(X_0=x)T(x,x)^{k-1}T(x,e)\\ &=&T(x,x)^{k-1}\sum_{e\in E\backslash\{x\}}T(x,e) \end{eqnarray*} The third equality comes from the fact that $\mathbb P_x(X_0=x_0,...,X_k=x_k)=\mathbb P_x(X_0=x_0)T(x_0,x_1)...T(x_{k-1},x_k)$ by definition of a transition matrix and we also have $\mathbb P_x(X_0=x)=1$.
Now, we have :
\begin{eqnarray*} \mathbb P_x(\tau=\infty)&=&1-\mathbb P_x\left(\bigcup_{k\in\mathbb N^*}\tau=k\right)\\ &=&1-\sum_{k\in\mathbb N^*}\mathbb P_x(\tau=k)\\ &=&1-\sum_{e\in E\backslash\{x\}}T(x,e)\sum_{k=0}^\infty T(x,x)^k\\ &=&1-(1-T(x,x))\frac{1}{1-T(x,x)}\quad\text{because $\sum_{e\in E}T(x,e)=1$}\\ &=&1-1=0 \end{eqnarray*} and so $\tau<\infty$ $\mathbb P_x-$a.s.