Let $S_n$ be a simple random walk in $D \subset\Bbb Z^d$ started at $x \in D$. Let $\tau =\inf\{n \geq 0 : X_n \notin D\}$.
I need to show that this quantity is finite $\forall y \in D$: $$ \sum_{k=0}^\infty \Bbb P_x(X_k=y,k< \tau)< \infty$$
First, what I want to do is to find a $c \in [0,1)$ such that $\Bbb P_x(X_k=y,k< \tau)<c^k$ for all but finitely many $k$.
I don't know if it is helpful but I know that $ \sum_{k=0}^\infty \Bbb P_x(X_k=y,k< \tau)=\sum_{k=0}^\infty \#\{\text{walks }x \to y \text{ in } k \text{ steps within } D\}\times\big(\frac{1}{2d}\big)^k$.
However, I cannot find some nice bound for the number of walks from $x$ to $y$ $\text{ in } k \text{ steps within } D$. I'd like it to be smaller than some $b^k$, with $b<2d$ to use the result of geometric sums. Can someone help a find a way to show that the sum is finite ?
If we assume that the path cannot go back, then in the first step we have $2d$ possibilities, and in the following steps we have $2d-1$ possibilities, so we have $2d(2d-1)^{k-1}$ path from x to y, now note that $$\frac{2d(2d-1)^{k-1}}{(2d)^{k}}=\left(\frac{2d-1}{2d}\right)^{k-1}$$ and $2d-1<2d$, and therefore the series $$\sum_{k=0}^{\infty} \left(\frac{2d-1}{2d}\right)^{k}<\infty$$