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Let $\forall n=0,1,2,\dots$, $\alpha_{n}(x)$ are POLYNOMIALS in $x$. Next, let for all $x\neq0$ the power series $$\sum_{n=0}^{\infty}\alpha_{n}(x)t^{n}$$ has positive radius of convergence. Can one say the radius of convergence of the series $$\sum_{n=0}^{\infty}\alpha_{n}(0)t^{n}$$ is also positive?

My quess is NO, but I have no example.

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As you suspect, there are easy counterexamples. Take any $\alpha_n$ with $\alpha_n(0) = n!$ and $\alpha_n(x) = 0$ for $|x| \geq 1/n$. Then for non-zero $x$ the series is a polynomial, and $\sum_{n=1}^\infty n!t^n$ diverges except at 0. –  Ben Barber Apr 17 '13 at 11:51
    
No. Consider series $\sum_{n=0}^\infty n!e^{-n^2x^2}t^n$. –  Andrew Apr 17 '13 at 11:51
    
Very nice example, thx! –  Twi Apr 17 '13 at 18:05
    
Let's make it a little bit more difficult. Suppose, in addition, $\alpha_{n}(x)$ is a POLYNOMIAL in $x$. –  Twi Apr 17 '13 at 18:07
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closed as off topic by Benoît Kloeckner, Andreas Blass, Bill Johnson, Pietro Majer, Felipe Voloch Apr 19 '13 at 3:35

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1 Answer

Still no: $$ \sum_{n=1}^{+\infty}n!\left(1-\frac{x^2}n\right)^{n^3}t^n. $$

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I was wondering if there are still counter-examples when the power series is assumed to be convergent for all complex values of $x\neq 0$. Using Cauchy's formula it appears at first that some control exists on $\alpha_n(0)$, although there is a swap in quantifiers that needs some looking carefully, which I don't have that much time to do right now... –  Loïc Teyssier Apr 19 '13 at 7:31
    
Nice! Thank you! –  Twi Apr 19 '13 at 8:52
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