Series convergence for $|x| < 1$ Define 
$$
f(x) = 1 + \sum_{k = 1}^{\infty} \frac{\beta(\beta - 1) ... (\beta - k + 1)}{k!} x^k
$$
I need to show that $f(x)$ converges if $|x| < 1$.
From the condition for $x$ and the term $x^k$ I think the idea is to use the power series test and combine it with a comparison test. So I need an upper bound (or supremum) for the coefficients $\frac{\beta(\beta - 1) ... (\beta - k + 1)}{k!}$. Is this the right direction and if yes could you give a hint on how to do this?
Edit: I tried writing the coefficients this way
$$
\beta \frac{\beta - 1}{1} \frac{\beta - 2}{2} \frac{\beta - 3}{3} ... \frac{\beta - k+1}{k-1}\frac{1}{k} = \beta\left(\beta - 1\right)\left(\frac{\beta}{2} - 1\right)\left(\frac{\beta}{3} - 1\right)\dotsb\left(\frac{\beta}{k-1} - 1\right)\frac{1}{k}
$$
This is somewhat revealing but I'm still not sure what to do exactly.
 A: So that this question might have an answer, let's follow up on copper.hat's suggestion:

Theorem (The Ratio Test): Let $\{a_k\}_{k\in\mathbb{N}} \subseteq \mathbb{R}$ be a sequence of real numbers (note that we could also take this to be a sequence of complex numbers, but this generality is not required here), and define
  $$ L := \lim_{k\to\infty} \left| \frac{a_{k+1}}{a_k} \right|. $$
  If $L < 1$, then the series $\sum a_k$ converges, and if $L > 1$ then the series $\sum a_k$ diverges.  If $L = 1$, then the test is inconclusive.

For the given series, we have
$$ a_k = \frac{\beta(\beta-1)\dotsb(\beta-k+1)}{k!} x^k. $$
Thus for any $k\in\mathbb{N}$, we have
\begin{align} \frac{a_{k+1}}{a_k}
&= \frac{\beta(\beta-1)\dotsb(\beta-(k+1)+1) x^{k+1} }{(k+1)!} \cdot \frac{k!}{\beta(\beta-1)\dotsb(\beta-k+1)} \\
&= \frac{\beta(\beta-1)\dotsb(\beta-k+1)(\beta-k)}{\beta(\beta-1)\dotsb(\beta-k+1)} \cdot \frac{k!}{(k+1)k!}\frac{x^{k+1}}{x^k} \\
&= \frac{\beta-k}{k+1} x.
\end{align}
Hence, taking limits, we have
$$
L
= \lim_{k\to\infty} \left| \frac{a_{k+1}}{a_k} \right|
= \lim_{k\to\infty} \left| \frac{\beta-k}{k+1} x \right|
= \lim_{k\to\infty} \left| \frac{k-\beta}{k+1} \right| |x|
= |x|.
$$
Observe that $L < 1$ whenever $|x| < 1$.  Therefore the series
$$ \sum_{k = 1}^{\infty} \frac{\beta(\beta - 1) ... (\beta - k + 1)}{k!} x^k $$
converges by the ratio test whenever $|x|<1$.
