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$a_n$ 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.

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2 Answers 2

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 $1$.

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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 $$\dfrac1R=\lim\sup\sqrt[n]{|a_n|}\geq 1, \ \text{if } a_n\in\mathbb Z \text{ with infinitely many nonzero terms}.$$

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