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Unfortunately the argument that I posted contains a gap at the end, but I cannot delete the accepted answer. The gap is explained at the end of the proof, where I also explain what is really proved.

Let $$f(z)=\sum_{n=0}^\infty b_nz_n/n!,\quad g(z)=\sum_{n=0}^\infty a_nz^n/n!.$$ Then your relation between $a_n$ and $b_n$ means $$g(z)=e^{-z}f(z).$$ Suppose that $a_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform $$F(z)=\sum_{n=0}^\infty b_n z^{-n-1}$$ is analytic outside of this disk $D$. Thus the power series $$F(1/z)=\sum_{n=0}^\infty b_nz^{n+1}$$ has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since your condition that $b_nb_{n+1}=0$ implies that the density of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have a singularity on every arc of the circle of convergence which is bigger than semi-circle.

This argument does not work when $F(1/z)$ has infinite radius of convergence. To exclude this case, one may assume that $|a_n|$ tends to zero with geometric speed, that is $|a_n|=O(\delta^n)$ for some $\delta\in(0,1)$. Then $0\not\in K$, and $F(1/z)$ is not entire.

The argument above also shows that under the stated conditions one can conclude that $b_n=O(\epsilon^n),\forall \epsilon>0$.

And of course, if the conditiopn $b_nb_{n+1}=0$ is replaced by the condition that $f$ is even or odd, then the conclusion is true as well.

Reference.

L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2 (Chap. 1 also contains all other facts used in this argument).

Unfortunately the argument that I originally posted contained a gap at the end. The gap is explained at the end of the proof, where I also state 3 partial results.

Let $$f(z)=\sum_{n=0}^\infty b_nz_n/n!,\quad g(z)=\sum_{n=0}^\infty a_nz^n/n!.$$ Then your relation between $a_n$ and $b_n$ means $$g(z)=e^{-z}f(z).$$ Suppose that $a_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform $$F(z)=\sum_{n=0}^\infty b_n z^{-n-1}$$ is analytic outside of this disk $D$. Thus the power series $$F(1/z)=\sum_{n=0}^\infty b_nz^{n+1}$$ has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since your condition that $b_nb_{n+1}=0$ implies that the density of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have a singularity on every arc of the circle of convergence which is bigger than semi-circle.

This argument does not work when $F(1/z)$ has infinite radius of convergence.

1. To exclude this case, one may assume that $|a_n|$ tends to zero with geometric speed, that is $|a_n|=O(\delta^n)$ for some $\delta\in(0,1)$. Then $0\not\in K$, and $F(1/z)$ is not entire.

2. And of course, if the condition $b_nb_{n+1}=0$ is replaced by the condition that $f$ is even or odd, then the conclusion is true as well.

3. The argument above also shows that under the stated conditions one can conclude that $b_n=O(\epsilon^n),\forall \epsilon>0$.

Thus my argument needs either a stronger assumption (1) or (2), or leads to a weaker conclusion (3).

Reference.

L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2 (Chap. 1 also contains all other facts used in this argument).

Unfortunately the argument that I posted contains a gap at the end, but I cannot delete the accepted answer. The gap is explained at the end of the "proof".

Let $$f(z)=\sum_{n=0}^\infty b_nz_n/n!,\quad g(z)=\sum_{n=0}^\infty a_nz^n/n!.$$ Then your relation between $a_n$ and $b_n$ means $$g(z)=e^{-z}f(z).$$ Suppose that $a_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform $$F(z)=\sum_{n=0}^\infty b_n z^{-n-1}$$ is analytic outside of this disk $D$. Thus the power series $$F(1/z)=\sum_{n=0}^\infty b_nz^{n+1}$$ has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since your condition that $b_nb_{n+1}=0$ implies that the density of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have a singularity on every arc of the circle of convergence which is bigger than semi-circle.

This argument does not work when $F(1/z)$ has infinite radius of convergence. To exclude this case, one may assume that $|a_n|$ tends to zero with geometric speed, that is $|a_n|=O(\delta^n)$ for some $\delta\in(0,1)$. Then $0\not\in K$, and $F(1/z)$ is not entire.

The argument above also shows that under the stated conditions one can conclude that $b_n=O(\epsilon^n),\forall \epsilon>0$.

And of course, if the conditiopn $b_nb_{n+1}=0$ is replaced by the condition that $f$ is even or odd, then the conclusion is true as well.

Reference.

L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2 (Chap. 1 also contains all other facts used in this argument).

Unfortunately the argument that I posted contains a gap at the end, but I cannot delete the accepted answer. The gap is explained at the end of the proof, where I also explain what is really proved.

Let $$f(z)=\sum_{n=0}^\infty b_nz_n/n!,\quad g(z)=\sum_{n=0}^\infty a_nz^n/n!.$$ Then your relation between $a_n$ and $b_n$ means $$g(z)=e^{-z}f(z).$$ Suppose that $a_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform $$F(z)=\sum_{n=0}^\infty b_n z^{-n-1}$$ is analytic outside of this disk $D$. Thus the power series $$F(1/z)=\sum_{n=0}^\infty b_nz^{n+1}$$ has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since your condition that $b_nb_{n+1}=0$ implies that the density of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have a singularity on every arc of the circle of convergence which is bigger than semi-circle.

This argument does not work when $F(1/z)$ has infinite radius of convergence. To exclude this case, one may assume that $|a_n|$ tends to zero with geometric speed, that is $|a_n|=O(\delta^n)$ for some $\delta\in(0,1)$. Then $0\not\in K$, and $F(1/z)$ is not entire.

The argument above also shows that under the stated conditions one can conclude that $b_n=O(\epsilon^n),\forall \epsilon>0$.

And of course, if the conditiopn $b_nb_{n+1}=0$ is replaced by the condition that $f$ is even or odd, then the conclusion is true as well.

Reference.

L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2 (Chap. 1 also contains all other facts used in this argument).

Unfortunately the argument that I posted contains a gap at the end, but I cannot delete the accepted answer. The gap is explained at the end of the "proof".

Let $$f(z)=\sum_{n=0}^\infty b_nz_n/n!,\quad g(z)=\sum_{n=0}^\infty a_nz^n/n!.$$ Then your relation between $a_n$ and $b_n$ means $$g(z)=e^{-z}f(z).$$ Suppose that $a_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform $$F(z)=\sum_{n=0}^\infty b_n z^{-n-1}$$ is analytic outside of this disk $D$. Thus the power series $$F(1/z)=\sum_{n=0}^\infty b_nz^{n+1}$$ has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since your condition that $b_nb_{n+1}=0$ implies that the density of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have a singularity on every arc of the circle of convergence which is bigger than semi-circle.

This argument does not work when $F(1/z)$ has infinite radius of convergence. To exclude this case, one may assume that $|a_n|$ tends to zero with geometric speed, that is $|a_n|=O(\delta^n)$ for some $\delta\in(0,1)$. Then $0\not\in K$, and $F(1/z)$ is not entire.

Reference.

L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2 (Chap. 1 also contains all other facts used in this argument).

Unfortunately the argument that I posted contains a gap at the end, but I cannot delete the accepted answer. The gap is explained at the end of the "proof".

Let $$f(z)=\sum_{n=0}^\infty b_nz_n/n!,\quad g(z)=\sum_{n=0}^\infty a_nz^n/n!.$$ Then your relation between $a_n$ and $b_n$ means $$g(z)=e^{-z}f(z).$$ Suppose that $a_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform $$F(z)=\sum_{n=0}^\infty b_n z^{-n-1}$$ is analytic outside of this disk $D$. Thus the power series $$F(1/z)=\sum_{n=0}^\infty b_nz^{n+1}$$ has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since your condition that $b_nb_{n+1}=0$ implies that the density of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have a singularity on every arc of the circle of convergence which is bigger than semi-circle.

This argument does not work when $F(1/z)$ has infinite radius of convergence. To exclude this case, one may assume that $|a_n|$ tends to zero with geometric speed, that is $|a_n|=O(\delta^n)$ for some $\delta\in(0,1)$. Then $0\not\in K$, and $F(1/z)$ is not entire.

The argument above also shows that under the stated conditions one can conclude that $b_n=O(\epsilon^n),\forall \epsilon>0$.

And of course, if the conditiopn $b_nb_{n+1}=0$ is replaced by the condition that $f$ is even or odd, then the conclusion is true as well.

Reference.

L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2 (Chap. 1 also contains all other facts used in this argument).