Why does the valence formula imply $M_k=0$ for $k <0$?

Question

I'm studying Modular Forms and in the notes I'm reading the author states the following result, known as the valence formula:

"Let $f$ be a non-zero weakly modular meromorphic form of weight $k$ on $SL_2(\mathbb{Z})$. Then:

$v_{\infty}(f)+\frac{v_i(f)}{2}+\frac{v_{\rho(f)}}{3}+\sum_{z \in \mathbb{H} / SL_2(\mathbb{Z})}^{*} v_z(f)=\frac{k}{12}$, where $\rho$ is the primitive cubic root of unity in $\mathbb{H}$"

Then, the author uses this result to prove that the set of modular forms of weight $k$ less than zero $M_k=\{0\}$ arguing the following:

Since the left-hand side of the valence formula is non-negative, the right hand side must be non-negative too, hence $k< 0 \implies M_k=\{0\}$.

My question is: why is the LHS of the valence formula non-negative?

Thank you in advance!

Answer 1

If $f \in M_k$ then $f$ is holomorphic everywhere (by definition), so $v_z(f) \ge 0$ for all $z$ (including $z = i, \rho, \infty$). So the LHS of the valence formula is a sum of non-negative quantities.