$f(\frac{1}{n})=a_n$ 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 ? 
 A: 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.
A: Suppose there is such a holomorphic function $f$ :
a)Consider the set $\{\frac 13, \frac 15 , \frac 17,...\} \subseteq D$. This set has $0$ as the limit point. and $f(\frac 1n)=\frac 1{n-1} \; \forall \; \text{odd} \; n$. Thus $f(z)=\frac {z}{1-z}$ on $D$ by Identity theorem. But then $f(\frac 12)=1 \neq 0$. A contradiction.
b)Consider the set $\{\frac 12, \frac 14, \frac 16,...\}$ which has $0$ as the limit point. Also $f(\frac 1n)=\frac 1n \; \forall \; \text{even} \;n$. Thus by Identity theorem $f(z)=z$ on $D$. But Again $f(\frac 13)=\frac 13 \neq \frac 12$. A contradiction.
c)Consider the subset $\{\frac 12, \frac 13, \frac 14, \frac 15, ...\}$ of $D$ which has $0$ as the limit point. Also $f(\frac 1n) = \frac {n-1}{n}=1-\frac 1n \; \forall \; n \in \Bbb N.$ Thus by Identity theorem $f(z)=1-z$ on $D$. Note that this is the required holomorphic function. So no contradiction in this case.
