I want to prove that if $f$ is a holomorphic function in a neighbourhood of $0$ and $|f(\frac{1}{n})| \le \frac{1}{e^n}$ for $n$ sufficiently big, then $f =0$.
I know that if $f$ is holomorphic in a neighb. of zero, then it has the form $f(\frac{1}{x}) = \sum _{n=0} ^{\infty} \frac{f^{(n)} (0)}{n! x^n}$
$$ \mathrm{e}^{-x} = 1-x + \frac{x^2}{2} - \frac{x^3}{6} +\cdots = \sum_{n=0} ^{\infty} \frac{(-1)^n x^n} {n!}$$
I thought I could use the identity principle which says that if $f,g \in \mathcal{O}(\Omega), \ f=g $ on $A \subset \Omega$ s.t. $A' \cap \Omega \neq \emptyset$, then $f=g$ on $\Omega$, but I have trouble determining the set $A$ in this case.
Could you help me a bit?
As $f$ is continuous at $z=0$, and $$ \left|\,\,f\left(\frac{1}{n}\right)\,\right|\le \frac{1}{\mathrm{e}^n}, $$ then $\,f(0)=\lim_{n\to\infty} f\big(\frac{1}{n}\big)=0$. If we assume that $f\not\equiv 0$, then $f$ is expressed as $$ f(z)=z^mg(z), $$ where $m$ is a positive integer and $g$ an entire function with $g(0)\ne 0$. But then we would have that $$ \left|\,\,f\left(\frac{1}{n}\right)\,\right|=\frac{1}{n^m}\left|\,g\left(\frac{1}{n}\right)\,\right|\le \frac{1}{\mathrm{e}^n}, $$ and hence $$ \left|\,g\left(\frac{1}{n}\right)\,\right|\le \frac{n^m}{\mathrm{e}^n}, $$ which means that $\,\lim_{n\to\infty} g\big(\frac{1}{n}\big)=0$, and as $g$ is continuous at $z=0$, then $g(0)=0$, which is a contradiction.
Thus $f\equiv 0$.