Hypergeometric function. Summation question

[LagrangeEuler]

Homework Statement

It is very well known that $\sum^{\infty}_{n=0}x^n=\frac{1}{1-x}$. How to show that
$\sum^{\infty}_{n=0}\frac{(a)_n}{n!}x^n=\frac{1}{(1-x)^a}$
Where $(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$
[/B]

Homework Equations

$\Gamma(x)=\int^{\infty}_0 e^{-t}t^{x-1}dt$

The Attempt at a Solution

$\frac{(a)_n}{n!}=\frac{a(a+1)...(a+n-1)}{n!}$
Not sure how to go further.[/B]

 

[geoffrey159]

Let's call for $\alpha < 0$ : $f_\alpha(x)= (1-x)^\alpha$ and $R_{n,\alpha}(x) = f_\alpha(x) - \sum_{k=0}^n \frac{f_\alpha ^ {(k)} (0)}{k!} x^k$

Can you explain why $|x| < 1$ and $x$ in a neighborhood of 0 implies $R_{n,\alpha}(x) = \int_0^x \frac{(x-t)^n}{n!} f_\alpha ^{(n+1)}(t) \ dt$. Then show that $\lim_{n\to \infty} R_{n,\alpha}(x) = 0$

 

[Ray Vickson]

Homework Statement

It is very well known that $\sum^{\infty}_{n=0}x^n=\frac{1}{1-x}$. How to show that
$\sum^{\infty}_{n=0}\frac{(a)_n}{n!}x^n=\frac{1}{(1-x)^a}$
Where $(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$
[/B]

Homework Equations

$\Gamma(x)=\int^{\infty}_0 e^{-t}t^{x-1}dt$

The Attempt at a Solution

$\frac{(a)_n}{n!}=\frac{a(a+1)...(a+n-1)}{n!}$
Not sure how to go further.[/B]

Just evaluate and simplify the $n$th term of the Maclaurin expansion of $f(x) = (1-x)^{-a}$.

Basically, you are being asked to verify that the binomial expansion of $(1-x)^n$ applies to non-integer and/or negative values of $n$. However, you cannot use the "factorial" definition of the binomial coefficient $C^n_k$ when $n$ is not a positive integer; instead, you need to use the explicit definition
C^n_k = \frac{n (n-1) \cdots (n-k+1)}{k!}
for integers $k \geq 0$.

 

[LagrangeEuler]

Thx a lot.