$\sum_\limits{n=1}^{\infty}\frac{nx}{(1+x)(1+2x)...(1+nx)}$ convergence domain.

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Problem: Study the absolute and conditional convergence domain of the series: $$\sum_\limits{n=1}^{\infty}\frac{nx}{(1+x)(1+2x)...(1+nx)}$$ 1) for $0\leqslant x\leqslant \epsilon$,$\epsilon>0$ and 2) $\epsilon\leqslant x<+\infty$.

I chose to apply the Alambert critera:$\lim_{n\to\infty}\frac{\frac{(n+1)x}{(1+x)(1+2x)...(1+(n+1)x)}}{\frac{nx}{(1+x)(1+2x)...(1+nx)}}=\lim{n\to\infty}\frac{nx+x}{nx+n^2x^2+nx^2}=\lim_{n\to\infty}=\frac{x+\frac{x}{n}}{x+nx^2+x^2}=0$

Question:

Is my attempt right? Because if it is. Why would the author ask the question for different convergence domain("1) for $0\leqslant x\leqslant \epsilon$,$\epsilon>0$ and 2) $\epsilon\leqslant x<+\infty$")?

Thanks in advance!