$f(\frac{1}{n})=a_n$ for which sequence holomorphic? Let $D=\{z\in\mathbb{C}\colon |z|<1\}$. Is there a holomorphic function $f\colon D\rightarrow\mathbb{C}$ with $f(\frac{1}{n})=a_n$ $(n=2,3,4,...)$ where $a_n$ is one of the sequences ?  
a) $a_n=0,\frac{1}{2},0,\frac{1}{4},0,\frac{1}{6},...$
b) $a_n=\frac{1}{2},\frac{1}{2},\frac{1}{4},\frac{1}{4},\frac{1}{6},\frac{1}{6},...$
c) $a_n=\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7},...$
Has anyone an idea on how to tackle this question ? 

If our function is holomorphic the derivatives at zero will all exist, and and any sequence of points approaching zero should give us the same derivative. So what happens for a)?
$\displaystyle\lim_{n \to \infty}\frac{f(\frac{1}{2(n+1)})-f(\frac{1}{2n})}{\frac{1}{2(n+1)}-\frac{1}{2n}}=\displaystyle\lim_{n \to \infty}\frac{\frac{1}{2(n+1)}-\frac{1}{2n}}{\frac{1}{2(n+1)}-\frac{1}{2n}}=1$, but $\displaystyle\lim_{n \to \infty}\frac{f(\frac{1}{2(n+1)+1})-f(\frac{1}{2n+1})}{\frac{1}{2(n+1)+1}-\frac{1}{2n+1}}=\displaystyle\lim_{n \to \infty}\displaystyle\frac{0}{\frac{1}{2(n+1)+1}-\frac{1}{2n+1}}=0$ which is a contradiction so no such function exists.
