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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?

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If $s$ is a nonnegative integer, things are easy... –  J. M. Jul 31 '12 at 11:18
    
@J.M. : He is using the notation for combination, so probably he means a non-negative integer? –  Jayesh Badwaik Jul 31 '12 at 11:18
    
@Jayesh, $\dbinom{n}{k}$ is the notation often used for any permissible arguments... –  J. M. Jul 31 '12 at 11:21
    
Sorry, i didn't mention that. $s$ can be an arbitrary complex number. –  Alex S Jul 31 '12 at 11:22
    
@J.M., Thanks for the info. –  Jayesh Badwaik Jul 31 '12 at 11:22

2 Answers 2

The limit is $0$ if $\def\Re{\operatorname{Re}}\Re s>-1$, it is nonzero and finite if $\Re s=-1$, and the sequence is unbounded if $\Re s<-1$. You see this by noting that $$\binom{s}{n}/\binom{s}{n-1}=\frac{s-n+1}{n}=-1+\frac{s+1}{n}$$ and considering the absolute value of this when $n$ is large, noting that the absolute value will be $$1-\frac{\Re s+1}{n}+O(n^2),$$ and using known facts about infinite products: $$\prod_{n=1}^\infty\Bigl(1+\frac{c}{n}\Bigr)=\begin{cases}0&c<0\\\infty&c>0\end{cases}$$ while $$\prod_{n=1}^\infty \Bigl(1+\frac{c}{n^2}\Bigr)$$ converges.

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The “known facts” about infinite products can be proved by taking logarithms, thus converting the products to sums. –  Harald Hanche-Olsen Jul 31 '12 at 11:47

If $s$ is a nonnegative integer, the answer is trivial, hence we assume that $s$ is not a nonnegative integer.

Let $x_n=(-1)^n\cdot{s\choose n+1}$, then $x_n\ne0$ for every $n$ and $\frac{x_n}{x_{n-1}}=1-\frac{s+1}n+O\left(\frac1{n^2}\right)$. Let $z_n=x_n\cdot\mathrm e^{(s+1)H_n}$ where $H_n=\sum\limits_{k=1}^n\frac1k$ is the $n$th harmonic number, then $\frac{z_n}{z_{n-1}}=1+O\left(\frac1{n^2}\right)$, hence $z_n\to z$ for a finite nonzero $z$. Thus, ${s\choose n+1}=(-1)^n\cdot\mathrm e^{-(s+1)H_n}\cdot(z+o(1))$.

Since $z\ne0$ and $H_n\to+\infty$, the sequence of general term ${s\choose n+1}$ converges if and only if ${s\choose n+1}\to0$ if and only if $\mathrm e^{-(s+1)H_n}\to0$ if and only if $\Re(s)\gt-1$.

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