Calculating infinite sum from a random walk in 1-dimension

Question

Consider the random walk on $\mathbb{Z}$ with 1-step transition probabilities $p_{i,j} = \frac{1}{2}$ iff $|i - j| = 1$. Then the probability that the first time that the chain returns from $i$ to $i$ is after exactly $2k$ steps, $f_{i,i}^{(2k)}$, is given by $$ f_{i,i}^{(2k)} = 2 C_{k-1} \left(\frac{1}{2} \right)^{2k}, $$ where $C_k$ is the $k^{th}$ Catalan number (https://en.wikipedia.org/wiki/Catalan_number), so $$ f_{i,i}^{(2k)} = \frac{2}{k} \binom{2k - 2}{k - 1} \left(\frac{1}{2} \right)^{2k}. $$ Each state is recurrent, so $$ \sum_{k = 1}^\infty f_{i, i}^{(2k)} = \sum_{k = 1}^\infty \frac{2}{k} \binom{2k - 2}{k - 1} \left(\frac{1}{2} \right)^{2k} = 1, $$ (and I have checked this sum on Wolfram Alpha), but I would be interested to see a proof of the sum equality. (I am not summing the odd terms $f_{i,i}^{(2k + 1)}$ as they are all equal to $0$.)

Answer 1

Let

$$f(x)=\sum_{k\ge 0}C_kx^k\;,$$

the generating function for the Catalan numbers; it’s well-known that

$$f(x)=\frac{1-\sqrt{1-4x}}{2x}\;.$$

Thus,

$$\sum_{k\ge 1}2C_{k-1}\left(\frac12\right)^{2k}=\frac12\sum_{k\ge 0}C_k\left(\frac14\right)^k=\frac12f\left(\frac14\right)=1\;.$$