binomial series convergence The series $\sum_{k=0}^{\infty} \binom{s}{k}, s\in\mathbb{R}   $ obviously evaluates to $2^s$ for $s\in \mathbb{N} $. But for general $ s\in \mathbb{R}$ the radius of convergence of   $\sum_{k=0}^{\infty}  \binom{s}{k}x^s $ such that $ \sum_{k=0}^{\infty}  \binom{s}{k}x^s= (1+x)^s$ is $1$, which means that $x = 1$ sits on the boundary of the disc of convergence. So now I'm not sure when $\sum_{k=0}^{\infty} \binom{s}{k}, s\in\mathbb{R}$ converges (to  $2^s$ by the Abel limit theorem). I think it converges for $s \in (-1,\infty)$ and diverges for $s\in (-\infty, -1]$. But I'm not sure, and I don't know how to prove it. Ideas?
 A: For $s \in \mathbb{R}\setminus \mathbb{N}$ and arbitrary $k \in \mathbb{N}$, we have
$$\binom{s}{k+1} = \frac{s-k}{k+1}\binom{s}{k}.$$
Hence for $s \leqslant -1$ we have
$$\biggl\lvert\binom{s}{k+1}\biggr\rvert \geqslant \biggl\lvert \binom{s}{k}\biggr\rvert,$$
so $\binom{s}{k}$ doesn't converge to $0$, and the series diverges.
For $s > -1$ the sequence of binomial coefficients is decreasing in absolute value, and eventually (once $k > s-1$) alternating, thus it suffices to see that $\binom{s}{k} \to 0$ for $s > -1$. Since
$$\log \biggl\lvert \frac{k-s}{k+1}\biggr\rvert = \log \biggl(1 - \frac{s+1}{k+1}\biggr) < - \frac{s+1}{k+1}$$
for $k > s$, and the harmonic series diverges, we find
$$\biggl\lvert\binom{s}{n}\biggr\rvert \leqslant \biggl\lvert \binom{s}{n_0}\biggr\rvert\cdot \exp \Biggl(-\sum_{k = n_0}^{n-1} \frac{s+1}{k+1}\Biggr) \xrightarrow{n\to\infty} 0$$
where $n_0 \geqslant s$. We further note that for $s > 0$ the convergence is absolute by Raabe's test.
A: And what is the behavior of $\sum \binom skx^k$ at $x=-1$? Raabe's test gives the answer easily, but here is an elementary approach. The terms
$$c_k:=\binom sk(-1)^k=\prod_{i=1}^k\frac{(i-1)-s}i$$ maintain the same sign when $k$ is sufficiently large. Let $k_0$ be any positive integer that exceeds $s+1$. For $k\ge k_0$ consider the product of positive factors
$$\left|\frac{c_k}{c_{k_0-1}}\right|=\prod_{i=k_0}^k\left( 1-\frac{s+1}i\right).\tag{$\ast$}
$$

Case $s>0$. Take the logarithm of both sides of $(*)$:
$$
\log \left|\frac{c_k}{c_{k_0-1}}\right|=\sum_{k_0}^k\log\left( 1-\frac{s+1}i\right)\stackrel{(1)}\le-(s+1)\sum_{k_0}^k\frac1i\stackrel{(2)}\le-(s+1)\log\frac k{k_0},$$ where (1) uses the inequality $\log x\le x-1$ and (2) uses $\sum_a^b\frac1i\ge\int_a^b\frac{dx}x$. Conclude
$$|c_k|\le |c_{k_0-1}|\left(\frac k{k_0}\right)^{-(s+1)}$$ and $\sum c_k$ converges absolutely by comparison with $\sum k^{-(s+1)}$.

Case $s<0$. Obtain a simple lower bound on $(\ast)$:
$$\left|\frac{c_k}{c_{k_0-1}}\right|=\prod_{i=k_0}^k\left( 1-\frac{s+1}i\right)> \prod_{i=k_0}^k\left( 1-\frac1i\right)=\frac{k_0-1}k,
$$
hence $\sum c_k$ diverges by comparison with the harmonic series.
