I would like to find the limit $ \lim_{n \to \infty} \binom{s}{n+1} = \lim_{n \to \infty} \frac{s (s-1) \cdots (s-n)}{(n+1)!} , $ where $s \in \mathbb C$. Actually, it would be enough to show that this sequence is bounded...
My aim is to show that for $x \in (-1, 1)$ we have $ (1+x)^s = \sum_{k=0}^\infty \binom s k x^k . $ So, by the Taylor theorem, we get for every $n \in \mathbb N$ $ (1+x)^s = \sum_{k=0}^n \binom s k x^k + R_n(f, 0)(x) . $ Now we need to show $ \lim_{n \to \infty} R_n(f,0)(x) = 0$ for every $x \in (-1,1)$. To do this, we first consider the case $x \in (0, 1)$. By the Lagrangian remainder formula we find for every $n \in \mathbb N$ a $\xi_n \in (0,x)$ such that $ | R_n(f,0)(x) | = \Bigg| \frac{ f^{n+1}(\xi_n) }{(n+1)!} x^{n+1} \Bigg| = \Bigg| \frac{s (s-1) \cdots (s-n) }{(n+1)!} \cdot (1 + \xi_n)^{s-n} \cdot x^{n+1} \Bigg| . $ My problem is now, that for $\text{Re}(s) < -1$ the term $\binom{s}{n+1}$ is not bounded... How can I solve this?