Does there exist an entire function $f$ such that $\forall n \in \mathbb{N}: f(\frac{1}{n}) = \frac{n}{2n-1}$

Question

I know that I am supposed to somehow apply the Identity Theorem, but other than that I am absolutely clueless.

Any hints?

Answer 1

Letting $n\to\infty$ gives $$f(0)=\frac12.$$ Also $$f(1/n)=\frac1{2-1/n}.$$ By the identity theorem, applied at $z=0$, $$f(z)=\frac1{2-z}$$ in a neighbourhood of zero. But this will continue to a function with a pole at $z=2$.

Answer 2

With $x=\frac{1}{n}$ we get $$f(x)=\frac{\frac{1}{x}}{\frac{2}{x}-1}=\frac{1}{2-x}$$