radius of symmetric random walk on $\mathbb{Z}$

Question

How to calculate the radius of the symmetric random walk on $\mathbb{Z}$, i.e. $\limsup_k (p^{(k)}(0,0))^\frac{1}{k}$?

($p^{(k)}(0,0)$ denotes the probability to get from $0$ to $0$ in $k$ steps and the transition probabilities of the symmetric random walk are $p(n,n+1)=p(n,n-1)=\frac{1}{2}$.)

Answer 1

When $k$ is odd, $p^{(k)}(0,0)$ is zero.

When $k$ is even, $p^{(k)}(0,0) = \frac{1}{2^k}\binom{k}{k/2}$. From Stirling's approximation, $$\lim_{k\to\infty } \binom{k}{k/2}^{1/k} =2$$ Hence, $\limsup_{k\to\infty } p^{(k)}(0,0)^{1/k} = 1$.