There are two similar questions in my complex analysis book that I can't for the life of me solve... They are about backing out a function given its Taylor Expansion. I will state both and give my attempt on both
(1) Find the holomorphic function whose Taylor Series expansion at $0$ is $$ \sum_{n=1}^{\infty} n^2 z^n \qquad \mbox{where } |z|<1 $$
The only obvious thing I can think of is to find a function $f$ such that $f^{(n)}(0) = n! n^2$, but really I am finding it impossible
(2) Find the holomorphic function whose Taylor Series expansion at $0$ is $$ \sum_{n=0}^{\infty} a_n z^n \qquad \mbox{where } |z|<1/2 \mbox{ and } a_n = \begin{cases} 2^n & n \mbox{ even} \\ 1 & n \mbox{ odd} \end{cases} $$
I can think to break the problem down in the following way: $$ f(z) = \left( \sum_{n=0}^{\infty} 2^{2n} z^{2n} \right) + \left( \sum_{n=0}^{\infty} z^{2n+1} \right) := g(z)+h(z)$$ Then I can find the function $g$ and $h$... But the problem is I can't see how to handle "Taylor Series" with the exponent being $2n$ and $2n+1$, how can I back out the original function?
I think the exercise is designed so that you use $\sum^\infty_{n=0}z^n = \frac{1}{1-z}=:h(z)$ together with the fact that $zh'(z)=\sum^\infty_{n=1} nz^n=z\frac{1}{(1-z)^2}$ and $z^2h''(z)=\sum^\infty_{n=2}(n^2-n)z^n=2z^2\frac{1}{(1-z)^2}$ in order to find the exact form of $\sum^\infty_{n=0}n^2z^n$ in terms of $\frac{1}{1-z}$. Try to work out the details from here.
Similarly, for the second function $g(z)$, you have \begin{align} g(z)&=\sum^\infty_{n=0}4^nz^{2n} + \sum^\infty_{n=0}z^{2n+1}\\ &=\sum^\infty_{n=0}(4z^2)^n+z\sum^\infty_{n=0}z^{2n}\\ &=\frac{1}{1-4z^2} + \frac{1}{1-z^2} \end{align}