Show that a certain entire function is identically zero Let $f(z)$ be an entire function satisfying $$ \left|\,f\left(\frac{1}{\log(n+2)}\right)\right|<\frac{1}{n}, \quad\text{for all $n\in \mathbb N$}$$ for $n\in \mathbb N$.
Show that $f(z)\equiv 0$.
I tried to use the uniqueness theorem but encounter a problem because the sequence 
$ f(z_n)=f\left(\dfrac{1}{\log(n+2)}\right)=0 $ 
only as $n\to \infty$ but to apply uniqueness theorem we need $f(z_n)=0$ for all $k$.
I stuck there.
Help wanted.
 A: We have $f(0)=0$ since $f$ is a continuous function in a neighbourhood of the origin and we have a sequence $\{x_n\}_{n\in\mathbb{N}}$ converging to $0$ such that $f(x_n)$ converges to zero. This gives that $f(x)=x\cdot g(x)$ where $g(x)$ is an entire function. But the same argument as above holds for $g(x)$...
A: Clearly, $f(0)=0$, and if $f$ is not identically zero, then there exists a positive integer $m$ and an entire function $g$, such that
$$
f(z)=z^m g(z),
$$
and $g(0)=0$. But in such case
$$
\left|\,f\left(\frac{1}{\log(n+2)}\right)\right|=
\left|\,\big(\log(n+2)\big)^{-m}g\left(\frac{1}{\log(n+2)}\right)\right|<\frac{1}{n},
$$
or
$$
\left|\,g\left(\frac{1}{\log(n+2)}\right)\right|\le\frac{\big(\log(n+2)\big)^{m}}{n}.
$$
But the left hand side of the above tends to zero, as $n\to\infty$, and hence so does the right hand side, and hence
$$
g(0)=\lim_{n\to \infty}g\left(\frac{1}{\log(n+2)}\right)=0,
$$
which is a contradiction.
Thus $f\equiv 0$.
Note. This is true even if we simply assume that $f$ is holomorphic is a neighborhood of zero.
A: This is only about the behavior at the origin. If $f\not \equiv 0$ then $f(x) = x^m \cdot g(x)$ with $g(0)\ne 0$. This implies 
$$\lim_{x\to 0} \frac{f(x)}{x^m}= \lim_{x\to 0} g(x) = g(0) \ne 0$$
and so 
$$\lim_{x\to 0} \frac{|f(x)|}{|x|^m}$$ 
is finite and nonzero for some $m$. 
Does not happen for the sequence
$$ \frac{|f (\frac{1}{\log(n+2)} )|} {|\frac{1}{\log(n+2)}|^m}$$
The sequence $|f (\frac{1}{\log(n+2)} )|$ decreases faster than any power of $|\frac{1}{\log(n+2)}|$since $(\frac{1}{n})$ does, too fast to be fit with a holomorphic function around $0$.
