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We have given a power series $\sum\limits_{n=0}^\infty a_nx^n$ and $a_0\ne0$. Suppose the power series has radius of convergence $R>0$. Then we may assume that $a_0=1.$

Explanation: It's $$\sum\limits_{n=0}^\infty a_nx^n=a_0\sum\limits_{n=0}^\infty \frac {a_n}{a_0}x^n$$

But why do $\sum\limits_{n=0}^\infty \frac {a_n}{a_0}x^n$ and $\sum\limits_{n=0}^\infty a_nx^n$ have the same radius of convergence $R$ ?

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Just go back to the definition of the radius of convergence of a power series: it's a number $R$ such that the series $\sum_{n=0}^{+\infty}a_nx^n$ is convergent if $|x|<R$. Since the convergence of $\sum_{n=0}^{+\infty}a_nx^n$ and $\frac 1{a_0}\sum_{n=0}^{+\infty}a_nx^n$ are equivalent, it gives the result.

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  • $\begingroup$ why are the convergence of $\sum_{n=0}^{+\infty}a_nx^n$ and $\frac 1{a_0}\sum_{n=0}^{+\infty}a_nx^n$ equivalent? $\endgroup$ – user31035 May 13 '12 at 14:38
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    $\begingroup$ It's just a result about sequences: if $\lambda$ is a real or complex number and the sequence $\{s_n\}$ converges, so does the sequence $\{\lambda s_n\}$. $\endgroup$ – Davide Giraudo May 13 '12 at 14:40
  • $\begingroup$ okay. thanks a lot! $\endgroup$ – user31035 May 13 '12 at 14:42
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    $\begingroup$ Well, if you are given only the radius of convergence, then you may assign any arbitrary values to the first finitely many coefficients. Because R= 1/(lim sup |a(n)|^(1/n)) and lim sup of a sequence does not depend on the first finitely many terms of it! $\endgroup$ – Somabha Mukherjee May 13 '12 at 15:17

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