prove absolute convergence $\sum_{n=0}^{\infty}\binom{x}{n}$ i am trying to prove that this series $\sum_{n=0}^{\infty}\binom{x}{n}$ is absolute convergent. let $\binom{x}{n}:=\frac{x^n}{n!}$
but i am stuck on the way, can someone please help me out. my steps using quotient test are: 
$$\frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n} = \frac{x^nxn!}{(n+1)!x^n} = \frac{1}{n!}\cdot \frac{x^nxn!}{(n+1)x^n}  = \ldots\text{help}\ldots = \theta < 1$$
i am trying to use the exponential series as a help since exponential series is absolute convergent. can you pls give me a hint. 
 A: Supposing $\binom{x}{n}:=\frac{x^n}{n!}$ (such a symbol should not! be used as it represents the binomial) we want to show
$$\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ is absolutely convergent (again supposing for $x\in \mathbb{R}$). By the ratio test,
$$\lim_{n\to +\infty}\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}=
\lim_{n\to +\infty}\frac{x^{n}n!}{(n+1)!}=\lim_{n\to +\infty}\frac{x^{n}}{n+1}=0<1$$
and we have the result. Observe that we don't have $x+1$ anywhere. $x$ is supposed to be a constant, only $n$ changes. You will want to note that
$$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$$ 
Now if
$$\binom{x}{n}=\frac{x(x-1)...(x-n+1)}{n!}$$
things are much more complicated. For $x=1$ the series converges while for $x=-1$ the series diverges. Indeed it converges only for $x>-1$. This page may be of interest.
Note: The fact that the last series converges only if $x>-1$, at least to my knowledge, requires the Taylor Theorem (The Integral Form of the Taylor Remainder). This is far from trivial to prove (if one doesn't know about Taylor series) but I will post it here if asked.
