$\sum n!a_n/(x(x+1)\cdots(x+n))$ and $\sum a_n/n^x$ has the same domain of convergence.

Question

I $\sum n!a_n/(x(x+1)\cdots(x+n)), x\neq 0,-1,-2,\cdots$ and II $\sum a_n/n^x$ has the same domain of convergence.

That is to say, if at $x$, $I$ converges, then $II$ converges. and vice versus.

Answer 1

Partial answer: For $x\geq 1$ $$ \frac{{n!}}{{x(x + 1) \cdots (x + n)}} = \Gamma (x)\frac{{\Gamma (n + 1)}}{{\Gamma (n + x + 1)}} \le \Gamma (x)\frac{1}{{n^x }} $$ (cf. http://dlmf.nist.gov/5.6.E8), and for $0<x<1$, $$ \frac{{n!}}{{x(x + 1) \cdots (x + n)}} = \Gamma (x)\frac{1}{{n^x }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right), $$ where the big-$\mathcal{O}$ is uniform in $x$ (cf. http://dlmf.nist.gov/5.11.E13). Hence if $II$ converges for $x\geq \sigma >0$ so does $I$.