I have the following question to solve. This is my attempt, please let me know if this is correct and if not, what can I do to further improve it. Thanks!
Question: Let $k \geq 4$ be even, let $\Gamma=\mathrm{SL}_2(\mathbb{Z})$ and take $\mathfrak{a}=\infty$. Recall the definition of the Eisenstein series $E_{\mathfrak{a}} \in M_k\left(\mathrm{SL}_2(\mathbb{Z}), \vartheta_{\mathrm{tr}}\right)$ from (32) and its Fourier expansion.
a) Evaluate the Fourier coefficients
$$ \eta_{\infty, \infty}(n)=\left(\frac{2 \pi}{i}\right)^{k} \frac{n^{k-1}}{\Gamma(k)} \sum_{c>0} c^{-k} S_{\infty, \infty}(0, n ; c) $$
as explicit as possible.
b) Conclude that the Fourier expansion of $E_{\mathfrak{a}}$ agrees with the one of the classical Eisenstein series $E_{k}$.
My attempted answer: (a) We are given the Fourier coefficient $\eta_{\infty, \infty}(n)$ as: $$ \eta_{\infty, \infty}(n)=\left(\frac{2 \pi}{i}\right)^k \frac{n^{k-1}}{\Gamma(k)} \sum_{c>0} c^{-k} S_{\infty, \infty}(0, n ; c), $$ where $S_{\infty, \infty}(0, n ; c)$ is the standard Kloosterman sum with $m=0$. The definition of the standard Kloosterman sum is: $$ S(m, n ; c)=\sum_{d \bmod c,(c, d)=1} e((\bar{d} m+d n) / c), $$ where $\bar{d}$ is the inverse of $d$ modulo $c$. When $m=0$, we are left with $S(0, n ; c)=\sum_{d \bmod c,(c, d)=1} e(d n / c)$. Since $e(z)$ evaluates to 1 when its argument is an integer, $S(0, n ; c)$ simplifies to the number of integers $d$ with $0<d<c$ and $\operatorname{gcd}(d, c)=1$, which is given by Euler's totient function $\varphi(c)$. We then substitute this into our expression for $\eta_{\infty, \infty}(n)$, yielding $$ \eta_{\infty, \infty}(n)=\left(\frac{2 \pi}{i}\right)^k \frac{n^{k-1}}{\Gamma(k)} \sum_{c>0} c^{-k} \varphi(c) $$ The sum is a Dirichlet series which for $\operatorname{Re}(\mathrm{k})>1$ converges to $\zeta(k) \zeta(k-1)$, where $\zeta(k)$ is the Riemann zeta function. Finally, we get the explicit form of the Fourier coefficients as: $$ \eta_{\infty, \infty}(n)=\left(\frac{2 \pi}{i}\right)^k \frac{n^{k-1}}{\Gamma(k)} \zeta(k) \zeta(k-1) $$
(b) We have that $E_{\mathfrak{a}}$ is a general form of the Eisenstein series which is defined, for $\mathfrak{a}=\infty$, as
$$E_k(\tau) = \frac{1}{2}\sum_{m,n\in\mathbb{Z}}' (\tau - m - n)^{-k},$$
where the prime on the sum indicates that the term $m = n = 0$ is to be omitted.
Let's compare this definition with the classical Eisenstein series $E_k$ which is defined as:
$$E_k(z) = 1 - \frac{4k}{B_k} \sum_{n=1}^\infty \sigma_{k-1}(n) q^n, $$
where $q = e^{2\pi iz}$, $\sigma_{k-1}(n) = \sum_{d|n}d^{k-1}$, and $B_k$ is the $k$th Bernoulli number.
From the definition of $E_{\mathfrak{a}}$, one can derive that it has the Fourier expansion:
$$E_{k}(\tau) = 1 - \frac{4k}{B_k} \sum_{n=1}^\infty \sigma_{k-1}(n) q^n$$
when restricted to $\tau \in \mathbb{H}$, where $\mathbb{H}$ is the upper half-plane. We see that this agrees exactly with the classical Eisenstein series $E_k(z)$.
Therefore, we can conclude that the Fourier expansion of $E_{\mathfrak{a}}$ agrees with the one of the classical Eisenstein series $E_k$, under the assumption that $E_{\mathfrak{a}}$ is a general form of the Eisenstein series as described above.
-- I would highly appreciate any suggestions/comments you have!