Show that if $ |f( \frac{1}{n}) | \leq \frac{1}{n!}$ then $0$ is an essential singularity

Question

Given holomorphic non-constant function $f:D(0,1) \smallsetminus \{0\} \rightarrow \mathbb{C}$ so $\forall n=2,3,...:\ |f(\frac{1}{n})| \leq \frac{1}{n!}$ I need do show that $0$ is an essential singularity.

I tried expanding $f$ to it's Laurent series near $z_0 = 0$ and compute coefficients by:

$$a_{-k} = \frac{1}{2 \pi i} \int_{|z|=r} z^{n-1}f(z)dz$$ and take $r= \frac{1}{n}$ so $$a_{-k} = \frac{1}{2 \pi i} \int_{|z|=\frac{1}{n}} z^{n-1}f(z)dz $$

But I don't know how to if it's a fertile direction nor how to proceed if it is.

Answer 1

Hint:

Does $$\lim_{z\to0}f(z)=L$$ exist? (For finite or infinite $L$)

Suppose it does, and because $1/n\to0$ then $|L|\leq0\to L=0$. Now, you must check that this value is possible.