The <Fourier expansion of a normalized Eisenstein series> is
$$
E_k(\tau)=1+\frac{2}{\zeta(1-k)}
\sum_{n\geq1}\sigma_{k-1}(n)q^n
=1-\frac{2k}{B_k}\sum_{n\geq1}\sigma_{k-1}(n)q^n,
\qquad q=e^{2\pi i\tau}.
$$
Thus
$$
a_0(E_k)=1,
\qquad
a_n(E_k)=\frac{2}{\zeta(1-k)}\sigma_{k-1}(n)quad(n\geq1).
$$
The Bernoulli number $B_k$ is rational, so every coefficient is rational. Equivalently, the standard Fourier calculation expresses the coefficient as a rational multiple of $(2\pi i)^k/\zeta(k)$, which is rational by the given fact that $\zeta(k)/\pi^k\in\mathbb Q$.

The supplied values give the familiar expansions
$$
E_4=1+240\sum_{n\geq1}\sigma_3(n)q^n,
\qquad
E_6=1-504\sum_{n\geq1}\sigma_5(n)q^n.
$$
Since every divisor sum is an integer, all their coefficients are integers.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 137 / 1 / a / Solution

For even $k\geq4$, the normalized level-one Eisenstein series has expansion
$$
E_k(\tau)=1+\frac{2}{\zeta(1-k)}\sum_{n\geq1}\sigma_{k-1}(n)q^n
=1-\frac{2k}{B_k}\sum_{n\geq1}\sigma_{k-1}(n)q^n.
$$


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