Which method i must use to this series (diverge or not) $$\sum_{n=1}^\infty~\frac{n!}{(a+1)(a+2)...(a+n)}$$
Maybe    $\lim_{n\to \infty} a_n \ne0$  ?
yes, sorry, $a>0$
 A: Since $a_n=\binom{a+n}{n}^{-1}$, try to represent it using the difference of two sequence elements when replacing $a$ by $a-1$ or $a+1$,
\begin{align}
\binom{a-1+n}{n}^{-1}-\binom{a+n}{n+1}^{-1}
&=
\frac{n!}{(a+n-1)...(a)(a-1)}-\frac{(n+1)!}{(a+n)...(a)(a-1)}\\[0.8em]
&=\frac{n!(a+n-(n+1))}{(a+n)...(a)(a-1)}=a_n
\end{align}
which allows you to examine this as telescoping series.
A: I just want to add another method. Not because it is going to provide a cleaner solution, but because the method itself could be useful for other problems. 
For dealing with the growth of factorials it is often convenient to use Stirling's 

$n!\sim n^ne^{-n}\sqrt{2\pi n}$ as $n\to\infty$.

Then the series is equivalent to the series of 
$$\frac{n^ne^{-n}\sqrt{n}a^ae^{-a}\sqrt{a}}{(a+n)^{a+n}e^{-(a+n)}\sqrt{a+n}}=\frac{n^na^a}{(a+n)^{a+n}}\frac{\sqrt{n}\sqrt{a}}{\sqrt{a+n}}$$
We can forget about $\frac{a^a\sqrt{n}\sqrt{a}}{\sqrt{a+n}}$, since it tends to a number different from zero.
We get 
$$\frac{1}{(a+n)^a}\frac{1}{(1+a/n)^n}.$$ 
Since the second fraction also converges to a number different from $0$ we can also forget about it.
Then the series is equivalent to $\frac{1}{n^a}$.
A: The difference of reciprocals of binomial polynomials of degree $a-1$ is proportional to the reciprocal of a binomial polynomial of degree $a$; to be precise,
$$
\frac1{\binom{n+a}{a}}=\frac{a}{a-1}\left[\frac1{\binom{n+a-1}{a-1}}-\frac1{\binom{n+a}{a-1}}\right]\tag{1}
$$
If we sum in $n$, the right side of $(1)$ telescopes. Thus, for $a\gt1$,
$$
\begin{align}
\sum_{n=1}^\infty\frac1{\binom{n+a}{a}}
&=\frac{a}{a-1}\frac1{\binom{a}{a-1}}\\
&=\frac1{a-1}\tag{2}
\end{align}
$$
For $a=1$, the series diverges since
$$
\sum_{n=1}^\infty\frac1{\binom{n+1}{1}}=\sum_{n=1}^\infty\frac1{n+1}\tag{3}
$$
diverges.
