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I want to check the behavior of $$\displaystyle\sum_{n=1}^\infty (z+\sqrt{5}+2i)^{n!}$$ outside its radius of convergence.

I've tried to use the ratio test as follows: $$\left|\frac{(z-\sqrt{5}+2i)^{(n+1)!}}{(z-\sqrt{5}+2i)^{n!}}\right|=|(z-\sqrt{5}+2i)^{nn!}|$$ This will converge to zero, if $|z-\sqrt{5}+2i|<1$.

However, what's about the cases $|z-\sqrt{5}+2i|=1$. How can I show convergence or divergence in this case?

EDIT: I've updated my question to make things more clear. Why I was down-voted for that?

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    $\begingroup$ There is no $z$ or other variable in your sum, therefore, you cannot ask for a radius of convergence. $\endgroup$ – Phira May 3 '14 at 19:54
  • $\begingroup$ @Phira - I'm sorry, that was a typo. The 1 should be z. $\endgroup$ – 0xbadf00d May 3 '14 at 20:00
  • $\begingroup$ possible duplicate of What is the radius of convergence of $\displaystyle\sum z^{n!}$? $\endgroup$ – Norbert May 3 '14 at 20:18
  • $\begingroup$ I disagree with the close votes. This question is about the boundary behaviour which is glossed over in the "duplicate" as "we know its behaviour". $\endgroup$ – Phira May 3 '14 at 20:33
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    $\begingroup$ @Phira - I've changed the question title to make things more clear. $\endgroup$ – 0xbadf00d May 4 '14 at 9:43
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Evidently, this is a power series around $-\sqrt 5 -2i$, so you really should make a change of variable and look at the power series $$\sum z^{n!}.$$

The ratio test for series will indeed give you the correct radius of convergence 1, as you have written in your question.

However, the ratio test never helps you for boundary behaviour.

On the boundary, you usually use one of the following approaches:

  1. terms do not even converge to zero, therefore divergent

  2. terms are positive for $z=R$ and converge (because of known series, integral test, etc), therefore absolute convergence on boundary

  3. terms converge for $z\not=R$ but on boundary due to Leibniz test/Abel criterion/Raabe test etc.

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  • $\begingroup$ Many of the Taylor coefficients are zero in the series $\sum z^{n!}$, hence the ratio test yields no information, since $\limsup ratio =\infty$ and $\liminf ratio = 0$. I would use the root test to conclude that the radius of convergence is one. $\endgroup$ – Bobby Ocean May 3 '14 at 21:14
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    $\begingroup$ @BobbyOcean This is not true. The ratio test does not only hold for power series, but for any series, and the OP clearly applied it to the series, not the power series which gives the correct radius of convergence. $\endgroup$ – Phira May 4 '14 at 9:11
  • $\begingroup$ If the ratio test is used as a test for convergence, then it might yield the radius of convergence as a consequence, however it can also fail and/or indicate incorrect information. For example, $$\frac{1}{2}+\frac{1}{3}z+\frac{1}{2^2}z^2+\frac{1}{3^2}z^3+\cdots,$$ has a radius of convergence, $\sqrt{2}$. In general, when one asks about the radius of convergence it is much more straight forward to use the formula that works every time, $$R=\left(\limsup \sqrt[n]{|a_n|}\right)^{-1}.$$ $\endgroup$ – Bobby Ocean May 4 '14 at 10:04
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    $\begingroup$ @BobbyOcean I am not opposed to using the limsup formula, I am opposed to your wrong claim that "the ratio test yields no information". It is perfectly reasonable to use it for this problem. $\endgroup$ – Phira May 4 '14 at 10:18
  • $\begingroup$ @oxbadfood If you read my comment, then you will see an example that shows your statement is incorrect. $\endgroup$ – Bobby Ocean May 4 '14 at 10:20
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If we wish to known the radius of convergence of $$\sum_{n=0}^\infty z^{n!}=z+z+z^2+z^6+z^{24}+\cdots, $$ then we calculate, $$\frac{1}{R} = \limsup \sqrt[n]{|\text{$n^{th}$ coefficient}|} = \limsup \sqrt[n]{\{0,2,1,0,0,0,1,0,0,0,\ldots\}}=1. $$ Rudin Principles of Mathematical Analysis - page 69 - Theorem 3.39

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  • $\begingroup$ Why is $|n^{th}\text{ coefficient}|=\left\{0,2,1,0,0,0,1,0,0,0,\ldots\right\}$? Why are so many values $0$? $\endgroup$ – 0xbadf00d May 3 '14 at 22:12
  • $\begingroup$ Unless I am misunderstanding the question, aren't we using a factorial as an exponent? Hence, the Taylor series (which doesn't use a factorial) looks like: $$\sum_{n=0}^\infty z^{n!} = 0z^0+2z^1+1z^2+0z^3+0z^4+0z^5+1z^6+0z^7+\cdots. $$ $\endgroup$ – Bobby Ocean May 3 '14 at 22:25
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    $\begingroup$ Your calculation gives the radius of convergence, but it tells us nothing about the behaviour on the boundary. $\endgroup$ – Phira May 4 '14 at 9:14
  • $\begingroup$ Your answer already discussed the boundary conditions? Which essentially says that each point on the boundary needs to be addressed individually. Was I suppose to repeat that? $\endgroup$ – Bobby Ocean May 4 '14 at 9:46

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