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I think if we have series $\sum_{n=1}^{\infty}a_nx^n$ and exist $a_n\neq0$ then series can have not more then countable number of roots, is it right? What theorem can proof it?

Sneach hcaens
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    Is the series absolutely convergent everywhere? Look at the uniqueness theorem in complex analysis. – copper.hat Apr 01 '20 at 05:36
  • @copper.hat, yes, we propose thate the series absolutely convergent everywhere. By using your hint: If the series has more then countable roots it means that exist the sequens of ${z_n}$ and $f(z_n)=0$ but we don't know about limit of $z_n$ – Sneach hcaens Apr 01 '20 at 05:44
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    Well, you can tile the plane with a countable number of squares. If there are an uncountable number of roots, one of those tiles will contain a countable number of roots and hence the roots have a limit point and so the function is identically zero. – copper.hat Apr 01 '20 at 05:47
  • Thank you, i got it! – Sneach hcaens Apr 01 '20 at 05:50

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If the series converges for all $x$ the the sum is an entire function. This implies that its zeros do not have any limit points. Hence the set of zeros is at most countable.

Kavi Rama Murthy
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