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Essentially I am trying to compute the radius of convergence for the following complex power series

$$\sum_{n=1}^\infty \frac{\ln n}{n^2}z^n$$

We have only been taught the methods of the root and ratio tests so far so I believe I am expected to use one of these.

I have attempted both but find them both to be inconclusive... my attempt at the root test is below.

$$\limsup_{n\rightarrow\infty} \sqrt[n]{\left| \frac{\ln n}{n^2} \right|} = \limsup_{n\rightarrow\infty} \frac{\sqrt[n]{\ln n}}{\sqrt[n]{n^2}} = \frac 1 1 = 1.$$ This is inconclusive.

I am sure there are easier methods to find this radius of convergence but within using the tests I have been taught I am not entirely sure if it is even possible or if perhaps I am missing something very obvious.

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    $\begingroup$ That looks quite conclusive to me. $\endgroup$ Oct 22 '17 at 18:06
  • $\begingroup$ @LordSharktheUnknown In what sense? I have been taught that when using the root test, if the limit is directly equal to 1 it provides no information. Is this incorrect? $\endgroup$
    – Evan
    Oct 22 '17 at 18:09
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    $\begingroup$ You get a limit of $1$, so the radius of convergence is $1$. $\endgroup$ Oct 22 '17 at 18:09
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Instead of $\displaystyle \limsup_{n\rightarrow\infty} \sqrt[n]{\left| \frac{\ln n}{n^2} \right|} = 1,$ you need $\displaystyle\limsup_{n\rightarrow\infty} \sqrt[n]{\left| \frac{\ln n}{n^2} z^n \right|} = |z|. $

Then you say the series converges if that is $<1$ and diverges if that is $>1.$

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    $\begingroup$ Well explained, clear, and concise. (+1) $\endgroup$
    – Mark Viola
    Oct 22 '17 at 18:16
  • $\begingroup$ @MarkViola : Thank you. $\endgroup$ Oct 22 '17 at 18:55
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The ratio test shows the series converges for $|z|<1$ and diverges for $|z|>1.$ The radius of convergence is therefore $1.$

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You're applying the root test to the series $\sum\frac{\ln n}{n^2}$, which is not the series you're trying to test for convergence. Apply the test to $\sum\frac{\ln n}{n^2}z^n$ instead and see what happens.

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