Let $f(z)$ be an entire function satisfying
$$\left|f\left(\frac{1}{\ln{(n+2)}}\right)\right|<\frac{1}{n}$$ for every $n\in\mathbb{N}).$ Show that $f(z)=0.$
I need some help for this question, I am looking for some hints which helps to solve this problem (not for possible solution).
I know pretty much popular theorems about entire functions, such as Liuville theorem, Casorati-Weierstrass and some nameless theorems. However, for this question, I am not sure what approach is the key.
If I am not mistaken $f\left(\frac{1}{\ln{(z+2)}}\right)$ has essential singularity at $z=-1$, so if I am right then maybe using Casorati-Weierstrass theorem is useful.
I can also say as $n\to\infty$, I get $|f(0)|\leq 0$, so $f(0)=0$. Hence, If I can show $f$ is constant it must be identically $0$.
By considering the limit as $n\to +\infty$ we have $f(0)=0$. If we assume $f(x)\not\equiv 0$ then $c_m x^m$, for some $m\in\mathbb{N}^+$ and $c_m\neq 0$, is the first non-zero term of the Maclaurin series of $f(x)$.
Let $M=\left|c_m\right|$ and $g(x)=\frac{f(x)}{x^m}$.
$g(x)$ is entire and in a neighbourhood $U$ of the origin we have $\left|g(x)\right|\geq \frac{M}{2}$.
On the other hand we also have
$$ \left|g\left(\frac{1}{\log(n+2)}\right)\right|\leq \frac{\log(n+2)^m}{n}, $$ so, no matter how small $M$ or $U$ are, for some large $n$ we get a contradiction.
It follows that $f(x)\equiv 0$.