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To find the radius of convergence of the series $\sum_{n=1}^\infty{n!(2x+1)^n}$ I tried the ratio test, $$\lim\limits_{n\rightarrow\infty}\left|\frac{(n+1)!(2x+1)^{n+1}}{n!(2x+1)^n}\right|=\lim\limits_{n\rightarrow\infty}\left|(n+1)(2x+1)\right|,$$ and if this $<1$ then the series converges. But this goes to infinity if $(2x+1)\neq 0$.

In this case is the radius of convergence is $0$?

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    $\begingroup$ Yes, it is zero. $\endgroup$ – Lord Shark the Unknown Aug 21 '17 at 6:15
  • $\begingroup$ Does the interval of convergence is zero then? $\endgroup$ – Xaviere Aug 21 '17 at 6:18
  • $\begingroup$ I think you have to find coeff. of $x^n$ first... $\endgroup$ – MAN-MADE Aug 21 '17 at 6:19
  • $\begingroup$ Also note that radius of convergence zero does not mean the domain of convergence is void. It does converge for one point... which one? $\endgroup$ – Miguel Aug 21 '17 at 6:33
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Your computations are O.K.

Consequence: the series $\sum_{n=1}^\infty{n!(2x+1)^n}$ converges $ \iff x=-1/2$.

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