Suppose I have a connected graph $G = (V,E)$, and a subset of vertices $U$, such that $W$ is the subgraph of $G$ induced by the vertices $V \setminus U$, and that $W$ is also connected.
Now, let $X_n$ be a random walk on $G$ and let $P$ be it's transition probability matrix. Now, let $Q$ be the submatrix of $P$ that corresponds to the vertices $W$. So, $Q$ is a sub-stochastic matrix (i.e. at least $1$ row sums to less than $1$)
I have shown that, for any $i,j \in W$, $\sum_{n=0}^\infty (Q^n)_{ij} = \mathscr{G}_U(i,j) < \infty$, and I have also shown that the power series $\sum_{n=0}^\infty x^n(Q^n)_{ij}$ converges for some $x > 1$, by showing that all eigenvalues of $Q$ are strictly less than $1$.
Now, I want to show that the radius of convergence, $R$, of $\sum_{n=0}^\infty x^n(Q^n)_{ij}$ does not depend on $i$ or $j$.
I am stuck here, and not sure how to proceed.
I'm pretty sure you're summing over $n$, not $i$.
Assuming that, from diagonalization you have that $(Q^n)_{ij}$ is some linear combination of $\lambda_k^n$ where $\lambda_k$ are all the eigenvalues of $Q$ (say sorted in decreasing order of magnitude). So the radius of convergence is presumably $1/|\lambda_1|$, except for the possibility that somehow $(Q^n)_{ij}$ had no contribution from $\lambda_1$. But the Perron-Frobenius theorem (for general nonnegative matrices, not just stochastic matrices) forbids that, because the eigenvector $v_1$ will have everywhere nonzero entries.