For any $z \in D(0,1) \subseteq \mathbb{C}$, the series identity $$(1-z)^{1/2} = \sum_{n=0}^\infty {1/2 \choose n} (-z)^n = 1-\sum_{n=1}^\infty \left|{1/2 \choose n}\right|z^n$$ holds. Letting $z \to 1-$ along the real axis, we deduce (aided by the monotone convergence theorem, or an ad hoc argument) that $$\sum_{n=1}^\infty \left|{1/2 \choose n}\right| = 1.$$ Can this identity be proven in an elementary (e.g. combinatorial) way?
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2The expression for $\binom{1/2}{n}$ in this answer seems like it may come in handy. – Semiclassical Dec 14 '19 at 00:17
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I'm afraid I can't follow the remark about the Monotone Convergence Theorem. I don't suppose you could expand on it a little? I'm sorry if it's something that should be obvious. Faced with the need to prove the same equation, I used Tauber's first theorem: Convergence of the sum of products $\sum_{k=0}^\infty \prod_{j=1}^k \left(1-\frac{3}{2j}\right)$. – Calum Gilhooley Sep 07 '20 at 17:57
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Letting $C_n=\frac1{n+1}\binom{2n}n$ be the $n^{th}$ Catalan number, then \begin{align} \sum_{n\ge 1}\left|\binom{1/2}n\right| &=\sum_{n\ge 1}(1/2)^{2n-1}\frac1n\binom{2n-2}{n-1} =\sum_{n\ge 1}(1/2)^{2n-1}C_{n-1} \end{align} Note $C_{n-1}$ is the number of simple random walk paths on the real line which stay to the right of the origin. The probability of such a path occurring is $(1/2)^{2n-2}$, because it has $2n-2$ steps. The last factor of $(1/2)$ can be interpreted as the probability of an additional left step. Therefore, the summand $(1/2)^{2n-1}C_{n-1}$ is the probability that the random walk hits $-1$ for the first time after $2n-1$ steps. Since such a random walk hits $-1$ eventually with probability one, these probabilities sum to one.
Mike Earnest
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