Show if there is an analytic function $f \in \mathbb{C}$ such that $f\left(\frac{1}{n}\right)=\frac{n}{1+n}$

Question

That's it. I don't really know what to do. I've tried to used analytic continuation with the function $g(z)=\frac{1}{1+z}$ but of course, such a function is not analytic in $\mathbb{C}$ so i can't use the theorem. And i can't find the contradiction

Answer 1

No such function exists that is analytic on all of $\mathbb{C}$. Indeed, suppose you have such a function $f$ and let $g(z)=\frac{1}{1+z}$. Then $f(1/n)=g(1/n)$ for all $n$, and also $f(0)=g(0)$ by continuity. So $f$ and $g$ are both analytic on the connected set $\mathbb{C}\setminus\{-1\}$ and agree on a non-discrete set, and so they must agree everywhere on $\mathbb{C}\setminus\{-1\}$. It follows that $f$ cannot be analytic (or even continuous) at $-1$.