Obviously, . It will be useful in some parts of the proof to note that, by symmetry, , so that Define Note first that . Moreover, for all , And since , we deduce that . This allows to define and for all and the previous identities implies Moreover, by the first entrance decomposition, Multiplying by and summing over yields . Finally, This is equivalent to so that . Only one of the roots has an expansion with only non-negative coefficients, that is, . This implies that that is, with probability 1, this random walk returns to 0.

Moroever, expanding yields: $A(z)=\sum_{k=1}^{\infty} \begin{pmatrix}2k-2\\k-1\end{pmatrix} \frac{1}{k 4^k} z^k\PP(\Delta=2k)=2\PP(\Delta=2k,S_1>0)=2\begin{pmatrix}2k-2\\k-1\end{pmatrix}\frac{1}{k 4^k}$.

Note that, setting such that , we have that are iid Bernoulli random variables with success probability . Then, where we have used the Stirling equivalence. This implies that where is the time index of the -th visit of to . By convention, we set . We now show by induction that for all , The case obviously holds. Now, assume that \eqref{eq:induc} holds for some . Then, and by the induction assumption, we get . Plugging \eqref{eq:induc} into \eqref{eq:one} yields Since , this implies that .