A holomorphic modular form is said to be a cusp form if its constant Fourier coefficients at the cusps are zero.
In the case of a modular form of weight $k$ for the full modular group $\mathrm{SL}_2(\mathbb Z)$, we can write
$$f(\tau) = F(q) = \sum_{n=0}^\infty a_nq^n,\quad q=e^{2\pi i \tau}, \tau \in \mathbb H$$
Then $f$ is a cusp form if $a_0=0$, because $\mathrm{SL}_2(\mathbb Z) \setminus \mathbb H$ has only one cusp, $\infty$.
According to the TASI lectures on moonshine, this is equivalent to $y^{k/2}f(x+iy)$ staying bounded as $y\rightarrow \infty$, but it seems that this would be true for any other exponent on $y$. What is special about $k/2$?