Determine a holomorphic function by means of its values on $ \mathbb{N} $ This is exercise 5, page 236 from Remmert, Theory of complex functions
For each of the following properties produce a function which is holomorphic in a neighborhood of $ 0 $ or prove that no such function exists:
i) $ f (\frac{1}{n}) = (-1)^{n}\frac{1}{n} \ $ for almost all $ n \in \mathbb{N}\ , n \neq 0 $
ii) $ f (\frac{1}{n}) = \frac{1}{n^{2} - 1 } $ for almost all $ n \in \mathbb{N}\ , n \neq 0, n \neq 1   $
iii) $|f^{(n)}(0)|\geq (n!)^{2}  $ for almost all $ n \in \mathbb{N} $  

Your title is misleading, as you cannot determine a holomorphic function from its values on $\mathbb{N}$. However, in this case you can determine it, using the uniqueness theorem for analytic functions: if $f$ and $g$ are two analytic functions and there is a convergent series $a_n$ such that $f(a_n)=g(a_n)$ for all $n$ then $f=g$.
Thus, for example, if your (i), putting $g(z)=z$. we see that $f(1/2n)=g(1/2n)$, so if $f$ is analytic we must have $f=g$. But then $f(1/(2n+1)) = 1/(2n+1)$, contradicting the fact that $f(1/(2n+1)) = -1/(2n+1)$, so no such analytic function exists.
(ii) is similar.
as for (iii), try to think on the Taylor expansion of $f$ near $0$. What will be its radius of convergence?
