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I have to find the radius of convergence of $\sum\limits_{n=1}^{\infty}n!x^{n!}$.

It is a power series, therefore:

One of the ways to find the radius of convergence is to find $\lim \dfrac{|n!|}{|(n+1)!|} = \lim \dfrac{n!}{(n+1)!} = \lim \dfrac{1}{n+1} = 0$

But in a lecture the professor simply computed $\lim \dfrac{(n+1)!x^{(n+1)!}}{n!x^{n!}} = \lim \ (n+1)x^{n!n}$ Therefore the radius of convergence is $1$.

Why are the results different and why does the second method work?

Did
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aribaldi
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    Only the second method work ; for a series $\sum u_n$, the interesting limit to compute is $\lim \frac{u_{n+1}}{u_n}$ – charmd Jun 06 '16 at 05:57
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    The second method is a correct application of the Ratio Test, which works for any series. The first method was done incorrectly; look closely at the statement you think you're using (it applies only when the coefficients are attached to all consecutive powers of $x$). – Greg Martin Jun 06 '16 at 07:01
  • @GregMartin What do you mean "when the coefficients are attached to all consecutive powers of x"? All I know is if you have a power series $\Sigma c_nx^n$ then the radius of convergence of that series is $lim sup |c_n|/|c_{n+1}|$ Why doesn't this apply in my case? – aribaldi Jun 06 '16 at 07:14
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    The method in your comment is obviously not the one you applied since here $c_n=n$ if $n$ is a factorial and $c_n=0$ otherwise hence $c_n/c_{n+1}$ is alternatively $0$ and $+\infty$... which cannot yield anything about the radius of convergence. – Did Jun 06 '16 at 07:32
  • @Did How the radius of convergence is $1$, in the second case? Isn't $n\to \infty$? – Kushal Bhuyan Jun 06 '16 at 07:53
  • "Second case" meaning second solution? @charMD answered that in their first comment, it seems, so what are you missing here? – Did Jun 06 '16 at 07:59
  • @Did "$\lim \dfrac{(n+1)!x^{(n+1)!}}{n!x^{n!}} = \lim \ (n+1)x^{n!n}$ Therefore the radius of convergence is $1$." How? Last limit is taken as $n\to \infty$, right? – Kushal Bhuyan Jun 06 '16 at 08:05
  • @KushalBhuyan One must complete this by noting that the limit in the RHS (yes, when $n\to\infty$, of course...) is zero when $|x|<1$ and is $+\infty$ when $|x|\geqslant1$ (eventhough $|x|>1$ would suffice). – Did Jun 06 '16 at 08:09
  • got it thanks @Did – Kushal Bhuyan Jun 06 '16 at 08:38
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    @aribaldi +1 for the clearly asked question. Are you in the position of collecting the indications given in the comments above and transforming them into a full solution? Then I suggest that you do so and post the resulting text as an answer. – Did Jun 06 '16 at 08:55

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