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I am facing the following problem.

Let $f$ and $g$ be analytic functions in $|z|<1$, with $$f(z)=\sum_{n=0}^{\infty}a_nz^n,\quad g(z)=\sum_{n=0}^{\infty}b_nz^n$$ such that $a_n\geq 0$, $b_n\geq 0$ and $f(z)g(z)=e^{z-1}$. Prove that $f$ and $g$ are entire functions.

After comparing the coefficient, I get $$a_nb_0+\cdots+a_0b_n=\frac{1}{n!e}.$$ I have tried to use this to prove that the radius of convergent is infinity, but I stuck with the expression above. What is a good approach to this question?

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    $\begingroup$ $0\leq a_n\leq \frac{1}{n!b_0e}$ $\endgroup$ – OR. Dec 4 '13 at 15:35
  • $\begingroup$ O I see, from this, I can conclude $\lim a_n=0$ and so $\lim\frac{a_{n+1}}{a_n}=0$ and hence radius of convergence is $\infty$. $\endgroup$ – Marco Dec 4 '13 at 15:44
  • $\begingroup$ From this you can conclude more. Compute what you want, the radius of convergence of $f$. $\endgroup$ – OR. Dec 4 '13 at 15:46
  • $\begingroup$ $\lim a_n=0$ doesn't imply $\lim a_{n+1}/a_n=0$. That is a wrong deduction. $\endgroup$ – OR. Dec 4 '13 at 15:50
  • $\begingroup$ Opps, I over look it, sorry. $\endgroup$ – Marco Dec 4 '13 at 15:51
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As ABC noted, due to the nonnegativity of coefficients, the identity $$a_nb_0+\cdots+a_0b_n=\frac{1}{n!e}\tag{1}$$ implies $$a_n \le \frac{1} {b_0 en!},\quad b_n \le \frac{1} {a_0 en!} \tag{2}$$ Neither $a_0$ nor $b_0$ can vanish, because $f(0)g(0)=e^{-1}\ne 0$.

The inequalities (2) imply that $\lim_{n\to\infty} a_n^{1/n}=0$ and $\lim_{n\to\infty} b_n^{1/n}=0$.

Caution: the inequalities (2) do not imply $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} =0$ or $\lim_{n\to\infty} \frac{b_{n+1}}{b_n}=0$

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