Is it true that for $\beta \in (0,1)$, we have:
$$\sum_{k=1}^{\infty} \left| C^{\beta}_{k} \right| = 1$$
where $C^{\beta}_{k}$ denotes the generalized binomial coefficient "beta choose k".
Sorry if this is something obvious. Haven't worked in this area before and it's appearing to me as an intermediary result for something else in another field.
Best Answer
I will use $\binom{\beta}k$ to denote $C^\beta_k$.
The Taylor series for $(1+x)^\beta$ around $x=0$ is$$(1+x)^\beta=\sum_{k=0}^\infty \binom{\beta}kx^k,\qquad -1<x<1,$$ The fact that this converges in the region $-1<x<1$ is easy to show using the ratio test. However, we need to show this converges when $x=-1$. Supposing we can show this, then your result follows: $$ 0=0^\beta=\sum_{k=0}^\infty \binom{\beta}k(-1)^k=1-\sum_{k=1}^\infty\left|\binom{\beta}k\right|, $$ where we have used $\left|\binom{\beta}k\right|=(-1)^{k-1}\binom{\beta}k$, valid for $k\ge 1$.
The series converges for $x\in (-1,1)$, and we want to show it converges at $x=-1$. Therefore, it suffices to show $$ \lim_{x\to (-1)^+}\sum_{k=0}^\infty \binom{\beta}kx^k=\sum_{k=0}^\infty \binom{\beta}k(-1)^k $$ To do this, we use the monotone convergence theorem for series. Writing this purported limit as $$ 1-\left(\lim_{x\to (-1)^+}\sum_{k=1}^\infty(-1)\binom{\beta}kx^{k}\right), $$ then all the summands, $(-1)\binom{\beta}kx^{k}$, are positive real numbers, and each summand increases as $x$ approaches $-1$ (from the right). We can then apply MCT to interchange the limit with the summation.