Does the series converge or diverge? I want to check, whether $$\sum\limits_{n=0}^{\infty }{\frac{n!}{(a+1)(a+2)...(a+n)}}$$
converges or diverges.
$a$ is a constant number
Ratio test
$$\begin{align}
  & \frac{a_{n}}{a_{n-1}}=\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}\cdot \frac{(a+1)(a+2)...(a+(n-1))}{(n-1)!}=\frac{n}{a+n} \\ 
 & \underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{a_{n-1}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{a+n}=1 \\ 
\end{align}$$
We know nothing
Root test
$$\sqrt[n]{a_{n}}=\sqrt[n]{\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}}=$$ I can't :(
I can only resolve the serie if "a" is a integer number
$$\begin{align}
  & \text{if }a\in \mathbb{Z} \\ 
 & (a+1)(a+2)...(a+(n-1))(a+n)=(n+a)(a+(n-1))...(a+2)(a+1)= \\ 
 & (n+a)(n+a-1)(n+a-2)...(a+2)(a+1)=\frac{(n+a)!}{a!} \\ 
 & \Rightarrow a_{n}=\frac{a!n!}{(n+a)!} \\ 
 & \text{if }a<0\Rightarrow \text{(}n+a)<n,a\in \mathbb{Z} \\ 
 & \underset{n\to \infty }{\mathop{\lim }}\,a_{n}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a!n!}{(n+a)!}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a!n(n-1)...(n+a)!}{(n+a)!}= \\ 
 & \underset{n\to \infty }{\mathop{\lim }}\,a!n(n-1)...(n+a-1)=\infty  \\ 
 & \Rightarrow \sum\limits_{n=0}^{\infty }{a_{n}=\infty } \\ 
 & \text{if }a>1,a\in \mathbb{Z} \\ 
 & a_{n}=\frac{a!n!}{(n+a)!}=\frac{a!n!}{(n+a)(n-1+a)...\underbrace{(n-a+a)!}_{n!}}=\frac{a!}{(n+a)(n-1+a)...(n+1)} \\ 
 & \Rightarrow  \\ 
 & \frac{a!}{\underbrace{(n+a)(n-1+a)...(n+1)}_{a\text{ times}}}<\frac{a!}{(n+1)^{a}}\Rightarrow  \\ 
 & \sum\limits_{n=0}^{\infty }{\frac{a!}{(n+a)(n-1+a)...(n+1)}<\sum\limits_{n=0}^{\infty }{\underbrace{\frac{a!}{(n+1)^{a}}<\infty }_{a>1}}} \\ 
 & \text{if }a=1 \\ 
 & \sum\limits_{n=0}^{\infty }{\frac{a!n!}{(n+a)!}}=\sum\limits_{n=0}^{\infty }{\frac{a!n!}{(n+1)!}}=\sum\limits_{n=0}^{\infty }{\frac{a!}{n+1}}=\infty  \\ 
\end{align}$$
But how i can resolve if "a" is not a integer number?
 A: Since:
$$(a+1)(a+2)\cdot\ldots\cdot(a+n)=\frac{\Gamma(a+n+1)}{\Gamma(a+1)}$$
assuming $\Re(a)>1$ we can write the original series as:
$$ S = \sum_{n\geq 1}\frac{\Gamma(n+1)\Gamma(a+1)}{\Gamma(a+n+1)}=\sum_{n\geq 1}n\cdot B(n,a+1)=\sum_{n\geq 1}n\int_{0}^{1}x^{n-1}(1-x)^a\,dx
\tag{1}$$
but $\sum_{n\geq 1}n x^{n-1}=\frac{1}{(1-x)^2}$ gives:
$$ S = \int_{0}^{1}(1-x)^{a-2}\,dx = \int_{0}^{1}x^{a-2}\,dx = \frac{1}{a-1}.$$
In order to prove that the condition $\Re(a)>1$ is necessary for the convergence of the series, just notice that the Euler product for the $\Gamma$ function gives:
$$\frac{\Gamma(n+1)}{\Gamma(n+a+1)}=\Theta\left(\frac{1}{n^a}\right)$$
hence the criterion for the convergence of the generalized harmonic series applies.
A: I don't know much about Gamma Function and summation under the integral so here is a more elementary proof.


*

*If $a>0$ we have $\ln a_n=-(\ln (1+\frac{a}{1})+\ln (1+\frac{a}{2})+\dots+\ln (1+\frac{a}{n}))$ and using


$$x\ge\ln(1+x)\ge x-\frac{x^2}{2}$$
(for $x>0$ proven by using monotonic function) we have
$$-a\sum_{i=1}^n\frac{1}{n}\le\ln a_n\le-a\sum_{i=1}^n\frac{1}{n}+\frac{a^2}{2}\sum_{i=1}^n\frac{1}{n^2}\le-a\sum_{i=1}^n\frac{1}{n}+C$$
(where $C=\frac{a^2}{2}\sum_{i=1}^\infty\frac{1}{n^2}$) Then using the fact that:
$$\ln(n+1)=\int_1^{n+1}\frac{1}{x}dx\leq\sum_{i=1}^n\frac{1}{n}\leq1+\int_1^{n}\frac{1}{x} dx=1+\ln(n)$$
we have
$$e^{-a(1+\ln n)}\le a_n\le e^{C-a(\ln(n+1))}$$
or
$$e^{-a}n^{-a}\le a_n\le e^C(n+1)^{-a}$$
Since $a_n$ is positive, the series converges if and only if $a>1$


*

*If $a\le 0$ then from $n\ge |a|$ we have $|a_n|$ increases and doesn't change sign so $\lim_{n\rightarrow\infty}\sup |a_n|\ne 0$ then the sum diverges.

