I am trying to solve the following problem using the maximum modulus principle:
I have a complex function $f$, analytic on an open and connected set $U$ containing $z = 0$, and I know that $|f(1/n)| < 1/2^n$ for all $n \in \mathbb{N}_{>0}$. I want to prove that this function is constant (and equal to zero) on $U$. Now I know that, by the maximum modulus principle, it would be sufficient to show that $|f(z)|$ attains a maximum on $U$ (it is not hard to show that $f(0) = 0$, so it then follows that $f = 0$ everywhere on $U$). However, I am not sure how to show this, and I am not even sure that this is the right way to do it. Any help or hints would be appreciated!
The maximum principle is not helpful here, because the information given concerns only one point $0$: it is a local issue. Locally, all we need to know about a holomorphic function is encoded in its power series $$ f(z) = \sum_{k=0}^\infty c_k z^k $$ The goal is to show $c_k=0$ for all $k$. Suppose this is not so. Let $m$ be the smallest index such that $c_m\ne 0$. Then $$ \frac{f(z)}{z^m} = \sum_{k=m}^\infty c_k z^{k-m} \to c_m \quad \text{as } z\to 0 $$ In particular, plugging $z=1/n$ here yields $f(1/n) n^m\to c_m$. But this contradicts the condition given, according to which $|f(1/n)|n^m < 2^{-n} n^m \to 0$.