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Show that the power series $\sum_1^\infty a_nz^n$ and $\sum_1^\infty na_nz^n$ have the same radius of convergence.

So I solved this problem using a method that I'm not sure is valid, and I just wanted to verify if this is rigorous enough and if not, try to understand the reason. Suppose $\sum_1^\infty a_nz^n$ has radius of convergence $R$, this implies $$\lim_{n\rightarrow \infty} |\frac{a_{n+1}}{a_n}|=\frac{1}{R}$$

Now using the ratio test for the second power series $$\lim_{n\rightarrow \infty} \left|\frac{(n+1)a_{n+1}}{na_n}\right|=\lim_{n\rightarrow \infty} \left|\left(1+\frac{1}{n}\right)\frac{a_{n+1}}{a_n}\right|=\frac{1}{R}$$

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    $\begingroup$ This is not valid. You cannot conclude that $\lim |\frac {a_{n+1}} {a_n}|=\frac 1 R$. This limit need not even exist. $\endgroup$ May 3, 2020 at 13:20
  • $\begingroup$ Oh I see, is there a simple example to when this occurs? $\endgroup$
    – Marcy
    May 3, 2020 at 13:25
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    $\begingroup$ $a_n=2$ for $n$ even and $a_n=1$ for $n$ odd gives an example. $\endgroup$ May 3, 2020 at 13:27

2 Answers 2

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Use the formula for radius of convergence and the fact that $\lim n^{1/n}=1$.

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  • $\begingroup$ What formula are you talking about here? $\endgroup$
    – Marcy
    May 3, 2020 at 13:32
  • $\begingroup$ @TigerAng The one derived from the root test, i.e. $R=\frac{1}{\lim \sup |c_n|^{1/n}}$. $\endgroup$
    – Botond
    May 3, 2020 at 14:00
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    $\begingroup$ $1/R=\lim \ sup |a_n|^{1/n}$. $\endgroup$ May 3, 2020 at 14:01
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Compare reciprocals of radii of convergence: note that$$\lim_{n\to\infty}n^{1/n}=1\implies\limsup_{n\to\infty}|na_n|^{1/n}=\limsup_{n\to\infty}|a_n|^{1/n}.$$

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