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Find the radius of convergence of the series

$$\sum\limits_{n=0}^\infty 3^nz^{n!}$$

My approach is as follow: $$R=\frac{1}{\lim\limits_{n\to\infty}\sup\sqrt[n]{a_n}}=\frac{1}{\lim\limits_{n\to\infty}\sup\sqrt[k!]{3^k}}=1.$$ Am I right? Any other approach would be apppriciated. Thanks in advanced.

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  • $\begingroup$ Yes, you're right $R=1.$ $\endgroup$ – user110661 Sep 26 '14 at 15:18
  • $\begingroup$ I think you're almost right. I don't see why you have a $3k!$ and not just $k!$ though. $\endgroup$ – amcalde Sep 26 '14 at 15:21
  • $\begingroup$ @amcalde: I made a mistake. I have edited it. $\endgroup$ – user3381651 Sep 26 '14 at 15:24
  • $\begingroup$ Indeed $R=1$ but the method is a little fast on the passage from $\sqrt[n]{a_n}$ to $\sqrt[k!]{a_k}$ (one might say a word about the other coefficients of the series, all the zeroes...). $\endgroup$ – Did Sep 26 '14 at 16:01
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OK, except I just noticed that you somehow switched to $k$ from $n$. Anyway since $\lim_{n\rightarrow \inf} 3^{n/(n!)} = \lim_{n\rightarrow \inf} 3^{-(n-1)!} = 1$

Your answer is correct.

As far as another approach, I would just use inspection: $|z^{n!}|$ grows incredably fast when ever $1 < |z|$ in order for the series to converge in that region, you would need coefficients that can keep it in check, at the least they need to bring the terms down to zero for even a chance at convergence. $3^n$ does not help, it only adds to the problem of divergence by making the terms yet larger in size.

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A different approach based on the quotient test (applied to the terms of the series, not to the coefficients). $$ \frac{3^{n+1}z^{(n+1)!}}{3^nz^{n!}}=3\,z^{(n+1)(n!-1)}. $$ As $n\to\infty$, this converges to $0$ if $|z|<1$ and to $\infty$ if $|z|>1$.

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  • $\begingroup$ Yes I was about to add this. Now I don't have to. Thanks! $\endgroup$ – amcalde Sep 26 '14 at 15:38

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