Suppose for a contradiction that $b_0\ne0$. The <Eisenstein series> in the question has constant term $2\zeta(k)$, so
$$
h=g-\frac{b_0}{2\zeta(k)}G_k
$$
has zero constant term and is therefore a level-one <cusp form> of weight $k$. Part (b) gives the coefficient bound $[q^n]h=O(n^{k/2})$. Since $b_n=O(n^{k/2})$ as well, the displayed Fourier expansion of $G_k$ would imply
$$
\sigma_{k-1}(n)=O(n^{k/2}).
$$

Take $n=\ell$ through the <prime number>[primes]. Then
$$
\sigma_{k-1}(\ell)=1+\ell^{k-1},
$$
which cannot be $O(\ell^{k/2})$ because $k-1>k/2$ for $k\geq4$. Therefore $b_0=0$, so $g$ vanishes at the only cusp of $\Gamma(1)$ and belongs to $S_k(\Gamma(1))$. This is the <Fourier coefficient growth criterion for a level-one cusp form>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 1 / c / Solution

Let even $k\geq4$ and let $g(\tau)=\sum_{n\geq0}b_nq^n$ be a level-one modular form of weight $k$. If $b_n=O(n^{k/2})$, then $b_0=0$ and $g$ is a cusp form. Otherwise subtract the matching Eisenstein series; its coefficients grow like $\sigma_{k-1}(n)$, contradicting the bound at prime indices.


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