Can a holomorphic function satisfy $f(1/n)=1/(n+1)$?

Solution 1:

Let $a_n:=\frac 1n$. Such a function should satisfy $f(a_n)=\frac{a_n}{1+a_n}$. Let $g(z):=f(z)-\frac z{z+1}$ on $B(0,1)$, the open ball. What can you say about the zeros of $g$?

If the question was

Does there exist a function $f$ which is holomorphic on $B_0( 1)$ (open disc of radius 1 in the complex plane) such that $f(1/n)=1/(n+1) \forall n \in \mathbb{N}$?

there would not be any contradiction, since $f(z):=z/(z+1)$ would do the job.

However, since we asked for the ball $B_0(2)$, the only function which can do the job is $z\mapsto z/(z+1$ for $|z|<1$, but this function cannot be continuous at $-1$, hence it cannot be holomorphic on the ball $B_0(2)$.