$$ \sum_{n=1}^\infty \frac{n! \> x^n}{(x+1)(2x+1) \cdots (nx+1)} $$
I start with d'Alembert, $\displaystyle L= \lim_{x\to\infty} \frac{a_{n+1}}{a_n}$ it gives me: $L=x+1$, to find convergence directly ($L<1$) we need $x<0$.
For $x>0$ the serie is not convergent because would be $L>1$.
Finally I try: $x=0 \Rightarrow L=1$. So There we have two ways: $$\frac{a_{n+1}}{a_n} \geq 1 \ or \ \frac{a_{n+1}}{a_n}<1$$
For $\displaystyle x=0 \Rightarrow \frac{a_{n+1}}{a_n}<1$ (The criteria does not decide, so I can apply Raabe:
I apply Raabe $\displaystyle (n(1-\frac{a_{n+1}}{a_n})$ i get the limit is $-\infty$ (It's less than 1), and as Raabe says the serie do not converges at $x=0$
I am right? Someone could check the solution with a math software or something, I do not know how to use matlab or mathematica.
Thanks in advance, Gabriel.
P.D. Sorry for the first edit, I am new at LaTeX and I did not have time enough to explain the exercise clearly.
How are you applying Rabee's criteria?
First of all, for $x=0$ clearly the series converges.
Now, for $x\neq 0$, $\dfrac{\frac{n!x^{n}}{(x+1)(2x+1)...(nx+1)}}{\frac{(n+1)!x^{n+1}}{(x+1)(2x+1)...((n+1)x+1)}}=\dfrac{(n+1)x+1}{(n+1)x}=1+\dfrac{1}{(n+1)x}$
If $x>0$, by Rabee's criteria, $n\left(\left(1+\dfrac{1}{(n+1)x}\right)-1\right)=\dfrac{n}{(n+1)x}\to \dfrac{1}{x}$. Then, if $0<x<1$, also series is convergent.
Analyse the case $x<0$.