For a real number $\alpha 0 $ call $\displaystyle \binom{\alpha}{n}:=\frac{\alpha(\alpha-1)(\alpha-2)...(\alpha+1-n)}{n!}$
the generalized binomial coeeficient. The products in the denominator and the numerator have $n$ factors
for each. Assume that $\alpha
otin \mathbb{N}$.\\
a) Prove: $\lim_{n o \infty}\frac{\binom{\alpha}{n+1}}{\binom{\alpha}{n}} = 1$

Question

For a real number $\alpha > 0 $ call $\displaystyle \binom{\alpha}{n}:=\frac{\alpha(\alpha-1)(\alpha-2)...(\alpha+1-n)}{n!}$
the generalized binomial coeeficient. The products in the denominator and the numerator have $n$ factors 
for each. Assume that $\alpha 
otin \mathbb{N}$.\\
a) Prove: $\lim_{n 	o \infty}\frac{\binom{\alpha}{n+1}}{\binom{\alpha}{n}} = 1$

paxpolaris · Answer

$$\displaystyle \binom{\alpha}{n}:=\frac{\alpha(\alpha-1)(\alpha-2)...(\alpha+1-n)}{n!}$$
Prove:
$$\Large \lim_{n \to \infty}\frac{\binom{\alpha}{n+1}}{\binom{\alpha}{n}} = 1$$

paxpolaris · Answer

$$\Large {{\alpha \choose n+1} \over {\alpha \choose n}}={a-n \over n+1}$$

anonymous · Answer

ls It answer?

paxpolaris · Answer

I just simplified it....

you still need to take the limit:
$$\Large \lim_{n \to \infty}{\alpha-n \over n+1}$$

paxpolaris · Answer

i don't know how it is =1 :'(

anonymous · Answer

thanks your effort tomorrow i must give very important homework i think i need a longer solution so that my mentor give me enough points :( and i have no idea how to start..

anonymous · Answer

$$(α(α−1)(α−2)...(α−n+1)(α−n)/(n+1)!)/(α(α−1)(α−2)...(α−n+1)/n!)$$
$$=(α−n)/(n+1)!)/1/n!=(α−n)n!/(n+1)!=(α−n)/(n+1)=α/(n+1)-n/(n+1)$$
When n goes to infinity, α/(n+1) goes to 0, and -n/(n+1) goes to -1.
So I get -1.

Let me try to do it again....

anonymous · Answer

ok it sounds hopefully

kinggeorge · Answer

I'm honestly not seeing how it could be something other than -1. If you take some number $\alpha$ and fix it in $\mathbb{R}^+ \setminus \mathbb{N}$, Then as $n$ goes to infinity, there comes a point where$$\binom{\alpha}{n}>0$$$$\binom{\alpha}{n+1}<0$$Because of the $\alpha\cdot(\alpha-1)\cdot...\cdot(\alpha-(n-1))$ terms.This means that the numerator and denominator of $$\Large {{\alpha \choose n+1} \over {\alpha \choose n}}$$will have differing signs when $\alpha-n<0$ and $\alpha-n+1>0$. Hence, as $n$ goes to infinity, it must be negative. And from what nightwill just showed above, it should be exactly $-1$.

kinggeorge · Answer

I was a little unclear at the end. They will have differing signs because if $\alpha\cdot(\alpha-1)\cdot...\cdot(\alpha-(n+1))$ has an odd number of negative numbers, then $\alpha\cdot(\alpha-1)\cdot...\cdot(\alpha-n)$ must have an even number of negative numbers and vice versa as 

$\displaystyle n\rightarrow \infty$

anonymous · Answer

Also KongGeorge are you agreed with the solution of Nightwill?

kinggeorge · Answer

I'm agreeing with nightwill

anonymous · Answer

ok thank you very much for your efforts

kinggeorge · Answer

Also, @PaxPolaris take a look at http://www.wolframalpha.com/input/?i=6.5+choose+9 The property you listed only seems to hold when $\alpha$ is a natural number.

paxpolaris · Answer

right sorry...