Power series expansion of an expresion

Question

I hope you can help me with a problem where I'm stuck. I need to expand $\frac{k!}{(1-st)^{k+1}}$ into $\sum_{n=0}^{\infty} \frac{(n+k)!}{n!}(st)^n$ and I don't know where to start. Thank's you in advance.

Answer 1

Hint:

If $f(x) = \dfrac{1}{1-x}$, then the $k^{th}$ derivative is $f^{(k)}(x) = \dfrac{k!}{(1-x)^{k+1}}.$ Now, use the geometric series to get a power series formula for $f(x)$ (valid of course only for $|x| < 1$).

Answer 2

Use the generalized binomial theorem:

$\begin{align*} (1 + x)^\alpha &= \sum_{n \ge 0} \binom{\alpha}{n} x^n \end{align*}$

Here:

$\begin{align*} \binom{\alpha}{n} &= \frac{\alpha^{\underline{n}}}{n!} \\ &= \frac{\alpha (\alpha -1) \dotsm (\alpha - n + 1)}{n!} \end{align*}$

If $\alpha$ is not a positive integer, the series converges only for $\lvert x \rvert < 1$.

If $\alpha$ is a negative integer, $\alpha = -k$:

$\begin{align*} &\frac{(-k)(-k - 1) \dotsm (-k - n + 1)}{n!} \\ &= (-1)^n \frac{k (k + 1) \dotsm (k + n - 1)} {n!} \\ &= (-1)^n \frac{(k - 1 + n)^{\underline{n}}} {n!} \\ &= (-1)^n \binom{n + k - 1}{n} \end{align*}$