Suppose $S_n$ is a symmetric random walk on the non-negative integers, and that $0$ is a reflecting state. Specifically, the transition probabilities are $p(0,1) = 1$ and $p(i,i\pm1) = \frac{1}{2}$. Let, $V_n$ denote the number of visits to state $0$ up until time $n$. Prove that for all $n \geq 1$,
$$ E_0[V_{n-1}] = E_0[S_n] - 1 $$
where the subscript means $0$ is the initial state.
We can write $V_{n-1}$ as a sum of indicator functions, $V_{n-1} = \sum\limits_{k=1}^{n-1}\chi_{\{S_k = 0\}} = \sum\limits_{k=0}^{n-1}\chi_{\{S_k = 0\}} - \chi_{\{S_0 = 0\}}$. Then, taking expectations,
$$ E_0[V_{n-1}] = \sum\limits_{k=0}^{n-1}P_0(S_k = 0) - \underbrace{P_0(S_0 = 0)}_{1} = \sum\limits_{k=0}^{n-1}P_0(S_k = 0) - 1$$
So it looks like we need to show that, $\sum\limits_{k=0}^{n-1}P_0(S_k = 0) = E_0[S_n] = \sum\limits_{k=0}^\infty kP_0(S_n = k)$. Some help would be greatly appreciated.
EDIT: In case the notation isn't familiar, the subscript implies a conditional probability, $E_0[V_n] = E[V_n | V_0 = 0]$
We can show the statement by induction on $n$. It can be checked by hand for $n =1$ and $n=2$.
Now assume $n>2$ and that the claim holds for all $ k < n$. Let's prove for $k = n$.
We can write $V_{n-1} = V_{n-2} + \chi_{ \{S_{n-1} = 0 \}}$, and hence $$ \mathbb{E} V_{n-1} = \mathbb{E} V_{n-2} + \mathbb{P}(S_{n-1} = 0) = \text{(apply the induction)} = \mathbb{E} S_{n-1} - 1 + \mathbb{P}(S_{n-1} = 0). $$ Thus, all we are left to show is $$ \mathbb{E} S_{n} = \mathbb{E} S_{n-1} + \mathbb{P} (S_{n-1} =0). $$
To prove the latter, and hence complete the induction step, observe that $$ S_{n} = S_{n-1} + X_{n}, $$ where $X_{n} = 1 $ if $S_{n-1} = 0$ and $\mathbb{P}(X_{n} = \pm 1 |S_{n-1} \neq 0) = \frac {1}{2}$. Hence, $$ \mathbb{E} S_{n} = \mathbb{E} S_{n-1} + \mathbb{P} (S_{n-1} =0) + \mathbb{E}(X_{n}|S_{n-1} \neq 0) . $$ But the last expectation is 0, since the event $S_{n-1} \neq 0$ is independent of $X_{n}$ (we have a random walk). Hence we are done.