Let $f:\mathbb C \to \mathbb C$ be an analytic function and $$\left|f\left(\frac{1}n\right)\right| \leq \frac{1}{n^{3/2}}$$ for each $n \in \mathbb N$. Now prove that the sequence $\left\{n^2f\left(\frac{1}n\right)\right\}$ is bounded.
Now notice that $$\left| f\left(\frac{1}n\right)\right| \leq \frac{1}{n^{3/2}} \implies\left|n^2f\left(\frac{1}n\right)\right| \leq \sqrt n$$ From here how can I solve this question?
It's a question about entire functions, but can be considered as a question about $C^2$ functions on a neighbourhood of $0$ in $\mathbb{R}$. Notice that the hypotheses imply $f(0) = f'(0) = 0$, so we have with l'Hospital $$\lim_{x \to 0} \frac{f(x)}{x^2} = \lim_{x\to 0} \frac{f'(x)}{2x} = \lim_{x\to 0} \frac{f''(x)}{2} = \frac{f''(0)}{2}$$