Let \(X_n\) be a symmetric random walk moving up or down by one at each step with equal probability. Let \(T\) be the waiting time until \(X_n=1\). Then \[
\def\E{\mathsf E}
\def\P{\mathsf P}
\begin{align}
\E[z^T]
&= \E[z \mid X_1=1]/2 + \E[z^{1+T' + T''} \mid X_1=-1]/2 \\
&= \frac{z + z\, \E[z^T]^2}{2}
\end{align}
\] where \(T'\) and \(T''\) are two independent copies of \(T\) that model the time to go from \(-1\) to \(0\) and then from \(0\) to \(1\). Solving the quadratic equation for \(\E[z^T]\) gives \[
\E[z^T] = \frac{1-\sqrt{1-z^2}}{z}.
\] Select the minus sign because the MGF must be \(<1\) when \(0<z<1\). This function is not differentiable at \(z=1\) and therefore \(T\) is not integrable (does not have a mean).
We can use the binomial formula to show that \[
\P(T=2n-1) = (-1)^{n-1}\binom{1/2}{n} = \frac{1}{2^{2n-1}(2n-1)}\binom{2n-1}{n}.
\]
