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I'm hoping someone might be able to verify my solution to the following problem:

Suppose that the series $\sum c_n z^n$ has radius of convergence $R$. Find the radius of convergence of the $\sum n^p c_n z^n$.

My solution: Let $(a_n)$ be the sequence of coefficients for the series. We have $(a_n) = (n^pc_n)_n$. Using the ratio test,

$$ L = \lim_{n\to\infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{n^p c_n}{(n+1)^pc_{n+1}} = \lim_{n\to\infty} \frac{n^p}{(n+1)^p} \cdot \lim_{n\to\infty} \frac{c_n}{c_{n+1}} = R \cdot \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^p = 1. $$

Thus, the radius of convergence is $1$.

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  • $\begingroup$ You cannot apply the ratio test in general. $\endgroup$ – Hagen von Eitzen Jul 30 '14 at 6:51
  • $\begingroup$ Wait why is that? $\endgroup$ – Jason Moor Jul 30 '14 at 6:52
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    $\begingroup$ There is no such theroem as "For any power series the raidus of convergence is given by $\lim\frac {a_n}{a_{n+1}}$." You cannot assume thet $\lim \frac{c_n}{c_{n+1}}$ exists - think of $c_n=1+(-1)^n$. Go for $\limsup\sqrt[n]{|a_n|}$. $\endgroup$ – Hagen von Eitzen Jul 30 '14 at 6:54
  • $\begingroup$ Ah, I see. But is the radius of convergence $1$? $\endgroup$ – Jason Moor Jul 30 '14 at 6:59
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    $\begingroup$ Now I'm getting $R$ for the radius of convergence, using the method suggested by Hagen. Is this correct? $\endgroup$ – Jason Moor Jul 30 '14 at 7:02
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$$ \limsup_{n\to\infty}\sqrt[n]{n^p|c_n|}=\left(\lim_{n\to\infty}\sqrt[n]{n}\right)^p\limsup_{n\to\infty}\sqrt[n]{|c_n|}=1\cdot\frac{1}{R}, $$ hence the radius of convergence is $R$.

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