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 $R$ ?


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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.