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If $a_n\neq 0$ for all $n \geq n_0$ and $\lim|\frac{b_n}{a_n}|=1$, then $\rho(S)=\rho(T)$. Since S=$\sum a_nz^n$ and T=$\sum b_nz^n$. I tried to use the definition of convergence radius

$$\limsup (|a_n|)^\frac{1}{n}=\frac{1}{\rho}$$

and I found that, if $\lim \frac{a_{n+1}}{a_n}$, $\lim\frac{b_{n+1}}{b_n}$ exists and for $n\geq n_0$ such that $|b_n |>0$, I have

$$\frac{b_n}{a_n}=\frac{\frac{a_{n+1}}{a_n}}{\frac{b_{n+1}}{b_n}}\frac{b_{n+1}}{a_{n+1}}.$$

Then

$$1=\lim \frac{b_n}{a_n}=\frac{\lim \frac{a_{n+1}}{a_n}}{\lim \frac{b_{n+1}}{b_n}}\lim \frac{b_{n+1}}{a_{n+1}}=\frac{\frac{1}{\rho (S)}}{\frac{1}{\rho (T)}}\left(\lim \frac{b_{n+1}}{a_{n+1}}\right)^{-1}=\frac{\rho (T)}{\rho (S)}.$$

Can I do that?

Thank you.

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  • $\begingroup$ I'm not sure I understand what you are asking... but it is true that $\lim \left|\frac{b_n}{a_n}\right|=\lim \left|\frac{b_{n+1}}{a_{n+1}}\right|=1$. In which case you can write $1=\frac{\lim \left|\frac{b_{n+1}}{a_{n+1}}\right|}{\lim\left|\frac{b_n}{a_n}\right|}$. $\endgroup$ – TravisJ May 8 '15 at 2:24
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The problem is that you do not know that $$ \Bigl|\frac{a_{n+1}}{a_{n}}\Bigr|\text{ and }\Bigl|\frac{b_{n+1}}{b_{n}}\Bigr| $$ converge.

Since $|b_n/a_n|\to1$, for $n$ large enough $$ \frac12\,|b_n|\le|a_n|\le2\,|b_n|. $$ From here it is easy to see that both series have the same radius of convergence.

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