Csdrhrt

Does there exist a nonconstant entire function with $f^{(n)}(0)=3^n$ for $n$ even and $f^{(n)}(0)=(n-1)!$ for $n$ odd.

My attempt:

If such $f$ exists, then $f(z)=sum_{n=0}^{infty}a_n,z^n$ where $$a_n=frac{f^{(n)}(0)}{n!}=begin{cases}frac{3^n}{n!}& n ,,text{even}\ frac{1}{n}& n,, text{odd}end{cases}$$

Then $f(z)$ can be rearranged such that begin{align*}f(z)&=sum_{n ,text{even}}frac{3^n}{n!}z^n+sum_{n ,text{odd}}frac{1}{n}z^n\&=sum_{k=0}^{infty} frac{3^{2k}}{(2k)!}z^{2k}+sum_{k=0 }^{infty}frac{1}{2k+1}z^{2k+1}\&=sum_{k=0}^{infty} frac{(9z^2)^{k}}{(2k)!}+sum_{k=0 }^{infty}frac{1}{2k+1}z^{2k+1}\&=cosh(3z)+tanh^{-1}zend{align*}

then $cosh(3z)$ is entire, but $tanh^{-1}z$ is analytic only when $|z|<1$. Therefore, $f(z)$ is not entire.

This question is from S. Ponnusamy and Herb Silverman book "Complex variables with applications ", and is considered in the exercise of section 8.2 which other parts solved by using the Identity theorem.

Is there another solution? Thanks in advance.

More simply: the radius of convergence of the Maclaurin series is $1$. For an entire function it would be $infty$.