$f(z)$ is analytic in the unit disk such that $|f(1/n)|\leq e^{-\sqrt{n}}$ for $n=2,3,...$ Prove that $f$ Is identically equal to zero.
My Attempt:
Consider the Taylor expansion $$f(z)=f(0)+f'(0)+\frac{f''(0)}{2!}+\frac{f'''(0)}{3!}+...$$By Cauchy estimate, for some $r=\frac{1}{n}$, we have $$\frac{|f^{(k)}|}{k!}\leq\frac{e^{-\sqrt{n}}}{(1/n)^k}$$
As $n\rightarrow \infty$, for each $k$, we have $\frac{|f^{(k)}|}{k!}=0$. Then, Taylor expansion suggests that $f(z)$ is identically zero.
I feel like I'm missing a lot of pieces. Any help or input is appreciated.
I don't know what you mean when you write “for some $r=\frac1n$”.
It follows from the squeeze theorem that $\lim_{n\to\infty}f\left(\frac1n\right)=0$. So, unless $f$ is the null function, the Taylor series of $f$ about $0$ will be of the type$$f(z)=a_kz^k+a_{k+1}z^{k+1}+\cdots$$with $k\in\mathbb N$ and $a_k\neq0$. So $\left(\left\lvert f\left(\frac1n\right)\right\rvert\right)_{n\in\mathbb N}$ behaves as $\frac{\lvert a_k\rvert}{n^k}$. But $\left(e^{-\sqrt n}\right)_{n\in\mathbb N}$ goes to $0$ much faster than that:$$\lim_{n\to\infty}\frac{\frac{\lvert a_k\rvert}{n^k}}{e^{-\sqrt n}}=\lvert a_k\rvert\lim_{n\to\infty}\frac{e^{\sqrt n}}{n^k}=\infty.$$