Let $S_n$ be any random walk. The following statement is usefully referred to as Spitzer's condition:
There exists $\rho \in (0,1)$ such that $$ \frac{1}{n} \sum_{k=1}^n P(S_n > 0 ) \to \rho. \qquad (1) $$
Now, it is said that the above is equivalent to $$\frac{1}{n} \sum_{k=1}^n P(S_n \geq 0 ) \to \rho, \qquad (2)$$ because $$\sum_{n=1}^\infty \frac{1}{n} P(S_n = 0) < \infty \qquad (3)$$ always holds true. I don't understand how (3) implies the equivalence of (1) and (2). I think I am missing something obvious here.
First notice that: $$\frac{1}{n} \sum_{k=1}^n P(S_k \geq 0 ) = \frac{1}{n} \sum_{k=1}^n P(S_k > 0 ) + \frac{1}{n} \sum_{k=1}^n P(S_k = 0 )$$ Thus, (1) and (2) are equivalent if: $$\frac{1}{n} \sum_{k=1}^n P(S_k = 0 ) \rightarrow 0$$ So, let's use (3) to prove this.
Equation (3) implies that for any $\epsilon > 0$ there exists a sufficiently large $M$ such that:
$$\sum_{k=M}^\infty \frac{1}{k} P(S_k = 0 ) < \epsilon/2$$
Then, we can find a $N$ sufficiently large that for all $n>N$:
$$\frac{1}{n} \sum_{k=1}^M P(S_k = 0 ) < \epsilon/2$$
Therefore, for all $n > N$:
$$\begin{align} \frac{1}{n} \sum_{k=1}^n P(S_k = 0 ) &< \frac{1}{n} \sum_{k=1}^M P(S_k = 0 ) + \sum_{k=M}^\infty \frac{1}{k} P(S_k = 0 ) \\ &< \epsilon \end{align}$$