The Problem: Consider a symmetric random walk started at zero. Show that with probability one it will return to zero. In other words show that if $T$ is the time of the first return to zero after the first step then $P(T<\infty)=1$. Here are two possible approaches.
$\mathbf{(1)}$ Use the fact that $P(T=2k)=\dfrac{1}{2k-1}\dbinom{2k}{k}\dfrac{1}{2^{2k}}$, and prove that $\displaystyle\sum_{k=1}^{\infty}P(T=2k)=1$ by writing down the Taylor expansion of an appropriate function.
$\mathbf{(2)}$ The random walk will be at $1$ or $-1$ after its first step. Thus it is enough to show that a random walk started at $1$ will always visit $0$. By symmetry the same is true if it starts from $-1$. For $k>0$ let $A_k$ be the event that the walk started from $1$ visits $0$ before $k$. Show that $(A_k)_k$ is an increasing sequence of events and that the union is exactly the event that the walk visits $0$ at some point.
My Thoughts: I will attempt the second approach as I have no idea how to go about the first one, even after writing out a few terms of the series, and I also don't see exactly how the identity would imply the result.
Let $A$ be the event that the random walk visits $0$ at some point and let $(A_k)_k$ be the sequence of events defined in $(2)$ above. Observe that if $A$ holds then the random walk visits $0$ at some point, say a fixed $n\in\mathbb N$, which means that the events $A_N$ hold for all $N>n$, and hence $A\subseteq\bigcup_{k\in\mathbb N}A_k$. For the reverse inclusion, note that if any $A_k$ holds, then the random walk visits $0$ at some point, so that $A_k\subset A$ for all $k\in\mathbb N$, and hence $\bigcup_{k\in\mathbb N}A_k\subseteq A$. To see that the sequence $(A_k)_k$ is increasing, observe that if $A_k$ holds, then the random walk visits $0$ before $k$ and hence also before $k+1$. It follows that $A_k\subseteq A_{k+1}$ holds for all $k\in\mathbb N$. Now the continuity of the probability measure implies that
$$P(A)=\lim\limits_{k\to\infty}P(A_k).$$
So we only need to find the probability of $A_k$ and show that it goes to $1$ as $k$ goes to infinity. To see this, we recall the gambler's ruin problem and note that
$$P(A_k)=1-\frac{1}{k},$$
since this is the event of leaving the game with no money before we reach $k$ dollars after starting with $1$ dollar. It now follows that
$$P(A)=\lim\limits_{k\to\infty}P(A_k)=\lim\limits_{k\to\infty}1-\frac{1}{k}=1.$$
Do you agree with my proof presented above? In addition, I would like to ask for help with the first approach suggested above, if possible.
Thanks a lot for your time, your help is much appreciated.
Note: I just realized that I have already asked about the second approach here Symmetric Simple Random Walk Eventually hits Zero, hence, I only have questions about how to go about the first approach.