The first time visiting a vertex is finite almost surely but its expectation is infinite

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Let $ (X_n)_{n \in \mathbb{N}} $ be a simple random walk on $ \mathbb{Z} $ starting from $1$. Let $ T_0 = \inf\{ n \geq 0 : X_n=0 \}$.

a) For an integer $N > 0$, compute the Green's function of the simple random walk $ (X_n)_{n \in \mathbb{N}} $ on $ \{ 0,\ldots,N\}$ with boundary $ \{ 0,N\}$.

b) Show that $ \mathbb{P}[T_0 < \infty ] = 1 $ and $ \mathbb{E}[T_0] = \infty $

Hint: compute $ \mathbb{E}[T_{0,N}] $ where $T_{0,N} = \inf \{ n \geq 0: X_n \in \{0,N\} \}$

For point a) I don't know how to continoue and for point b) I don't know for the second part.

For a) The Green's function, denoted $G(x,y)$ is the unique solution of $$ \left\{\begin{matrix} \Delta f(x) = - \delta_{x=y} &\text{if } x \in \{1,\ldots,N-1\} \\ f(x)=0& \text{if } x \in \{0,N\} \end{matrix}\right.$$ and where $y \in \{1,\ldots, N-1\} $. and here $ \Delta f(x) = \frac{f(x-1)+f(x+1)}{2} - f(x) $. Hence we have that when $x=y$ we found that $$ -1 = \frac{f(y-1)+f(y+1)}{2} - f(y) $$ and since $f(0)=f(N)=0$ then at $y$ we have a left slope of $f(y)/y $ and a right slope of $-f(y)/(n-y)$. By the condition that $\Delta f(y) = -1$ we have that $$ 1 = \frac{1}{2} \left( \frac{f(y)}{y} + \frac{f(y)}{N-y} \right) $$ and we found that $$ f(y) = \frac{2 y(N-y)}{N} $$ Hence $ G(0,y)=G(n,y) = 0 $ and $ G(y,y) = \frac{2 y(N-y)}{N} $, but I don't know how to compute $G(x,y) $ when $ x \not\in \{ 0,y,N\} $.

For point b) First of all since the simple random walks on $\mathbb{Z} $ starting from $1$ is recurrent then we have for any other vertex, in particular for $0$, that $\mathbb{P}_1(\exists n, X_n= 0) = 1$ hence we have that $\mathbb{P}(T_0 < \infty ) = 1$. For the second part I think that we have to set $T_{0,N} = \inf \{ n \geq 0: X_n \in \{0,N\} \} $ and we have that $$ G(x,y) = \mathbb{E}_x \left( \sum_{n=0}^{T_{0,N}-1} \delta_{X_n=y} \right) $$ I want to prove that $G(x,y) < \mathbb{E}[T_{0,N}] $ and then proving that when $N \to \infty $ we have that $G(x,y) \to \infty $ and then deducing the result from the fact that when $N \to \infty $ we have that $T_{0,N} \to T_0$