How do I show that a holomorphic function which satisfies this bound on reciprocals of integers is identically zero?

Question

I'm trying to show that $f$ is a holomorphic (around the origin) map satisfying $|f(\frac{1}{n})|\leq \frac{1}{2^n}\forall n\in\mathbb N$ then $f$ is identically zero. Usually, if some function equals an analytic function on a dense set, we can invoke uniqueness of analytic continuation. But in this case, we don't have equality, just a bound. I tried Cauchy's integral formula also, but to no avail. How do I solve this problem? Thank you.

Answer 1

If $f$ is not identically zero there is a $p\geq0$ and an analytic function $z\mapsto g(z)$ with $g(0)=:c\ne0$ such that $$f(z)=z^p g(z)\qquad\bigl(z\in U(0)\bigr)\ .$$ It follows that $$2^n f\left({1\over n}\right)={2^n\over n^p}\>g\left({1\over n}\right)\to\infty\qquad(n\to\infty)\ ,$$ contrary to assumption.