I was studying for my Calculus final test and I was struggling studying the convergence of the following series:
For $\alpha >0$, $ \sum_{n=0}^{\infty} \frac{n!}{(a+1)...(a+n)}$
I first applied D'Alemberts test, but $\lim_{n \to \infty} (x_{n+1}/x_n)=\lim_{n \to \infty} (\frac{n+1}{n+\alpha + 1})= 1$. The test doesn't give any information, but from the equality $x_{n+1}/x_{n} = \frac{n+1}{n+\alpha + 1}$, as $\alpha > 0$, then $n+1 < n+\alpha+1$; thus $\frac{n+1}{n+\alpha + 1} < 1$ and we conclude that $x_{n+1}<x_n$ hence the sequence $x_n$ is decreasing.
The other possible approach I though of was using Raabe's test; $\lim_{n \to \infty}(n \cdot(1-x_{n+1}/x_n))=\lim_{n \to \infty}(\frac{\alpha n}{\alpha + n +1})= \alpha$ Thus for $\alpha > 1$ the series converges and $\alpha < 1$ the series diverges.
However, I have two issues. First of all, we haven't studied Raabe's test in our Calculus class (in the exam we are only allowed to use comparison tests, root and D'Alembert's test and Leibniz's theorem); hence I would like to know if there's an alternative way of doing this exercise without using Raabe's test. Moreover, how can I determine whether if the series converges for $\alpha = 1$? I would also appreciate any correction to my solution as I'm not familiar with Raabe's test.
