Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$a_n$ be a sequence of integers such that such that infinitely many terms are non zero, we need to show that either the power series $\sum a_n x^n$ converges for all $x$ or Radius of convergence is at most $1$. need some hint. thank you.

share|cite|improve this question
up vote 4 down vote accepted

Claim: let $\{a_n\}$ a sequence of integers. The radius of convergence of $\sum_{n=1}^{+\infty}a_nx^n$ is

  • infinite if $a_n=0$ when $n$ is large enough;
  • at most $1$ if infinitely many $a_n$ are different from $0$.

The first case is trivial.

If $|z|>1$ and $n$ is such that $a_n\neq 0$, as $a_n$ is integer we have that $|a_n|=1$ so $|a_nz^n|\geqslant |z|^n$. As it occurs for infinitely many terms, this proves that the sequence $\{a_nz^n\}$ is not bounded whenever $|z|\geqslant 1$, hence the radius of convergence is at most $1$.

share|cite|improve this answer
Now the question is: in the second case, which numbers can be obtained as a radius of convergence of such series? For example, with $a_n:=p^n$, we get numbers of the form $1/p$. – Davide Giraudo Jan 14 '13 at 14:20

The Radius of convergence, $R$, is given by $$\dfrac1R=\lim\sup\sqrt[n]{|a_n|}\geq 1, \ \text{if } a_n\in\mathbb Z \text{ with infinitely many nonzero terms}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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