I played around with the Fourier series of the Eisenstein series resp. divisor sums and did some calculations, see below. Although the deduction is * not rigorous / wrong* (as the power series for the Bernoulli numbers is not convergent), I wonder, why the final result - at least in the area $0<Im(z)\leq \frac{1}{2}$ near the imaginary axis - is still so good? (Even the correction term in case of $E_2$, i.e. for $m=1$, is matched!).

\begin{equation} \boxed{\sum_{n=1}^{\infty}{\sigma_m(n)\,e(nz)}\approx\frac{B_{m+1}}{2(m+1)}\left(1-\frac{1}{z^{m+1}}\right)+\left[m=1\right]\frac{1}{2}\frac{1}{2\pi iz}} \end{equation}

*Can someone give a reason for this behavior, or can someone calculate the error term explicitly?*

For odd $m\in\mathbb{N}$: \begin{equation*} \begin{aligned} \sum_{n=1}^{\infty}{\sigma_m(n)\,e(nz)}&=\sum_{n=1}^{\infty}{\left(\zeta(m+1)\,n^m\sum_{q=1}^{\infty}{\frac{c_q(n)}{q^{m+1}}}\right)e(nz)}\\ &=\zeta(m+1)\sum_{q=1}^{\infty}{\frac{1}{q^{m+1}}\sum_{n=1}^{\infty}{\sideset{}{^\star}\sum\limits_{a\leq q}{n^m\,e\left(z+\frac{a}{q}\right)^n}}}\\ &=\zeta(m+1)\sum_{q=1}^{\infty}{\frac{1}{q^{m+1}}\sideset{}{^\star}\sum\limits_{a\leq q}\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\left(\frac{1}{e\left(z+\frac{a}{q}\right)-1}\right)}\\ &=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum_{q=1}^{\infty}{\frac{1}{q^{m+1}}\frac{1}{2\pi iz}\sum_{d|q}{\mu\left(\frac{q}{d}\right)}\frac{2\pi i dz}{e(dz)-1}}\\ &=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum_{q=1}^{\infty}{\frac{1}{q^{m+1}}\frac{1}{2\pi iz}\sum_{n=0}^{\infty}{\frac{B_n}{n!}\left(2\pi i z\right)^n \sum_{d|q}{\mu\left(\frac{q}{d}\right)d^n}}}\\ &=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum_{q=1}^{\infty}{\frac{1}{q^{m+1}}\frac{1}{2\pi iz}\sum_{n=0}^{\infty}{\frac{B_n}{n!}\left(2\pi i z\right)^n J_n(q)}}\\ &=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum_{n=0}^{\infty}{\frac{B_n}{n!}\left(2\pi i z\right)^{n-1}\sum_{q=1}^{\infty}{\frac{J_n(q)}{q^{m+1}}}}\\ &=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum_{n=0}^{\infty}{\frac{B_n}{n!}\left(2\pi i z\right)^{n-1}\frac{\zeta(m+1-n)}{\zeta(m+1)}}\\ &=-\sum_{n=0}^{\infty}{\left(\frac{B_n}{n!}\left(2\pi i z\right)^{n-1-m}\zeta(m+1-n)\prod_{k\leq m}{\left(n-k\right)}\right)}\\ &=\frac{m!}{\left(2\pi iz\right)^{m+1}}\zeta(m+1)-\frac{B_{m+1}}{(m+1)!}\zeta(0)m!+\left[m=1\right]\frac{1}{2}\frac{1}{2\pi iz}\\ &=\frac{B_{m+1}}{2(m+1)}\left(1-\frac{1}{z^{m+1}}\right)+\left[m=1\right]\frac{1}{2}\frac{1}{2\pi iz} \end{aligned} \end{equation*}

Notation: $c_q(n)$ is Ramanujan's sum, $J_n(q)$ Jordan's totient function, $B_n$ the Bernoulli numbers, $\sideset{}{^\star}\sum\limits_{a\leq q}{}$ means $\sum\limits_{\substack{a\leq q\\\gcd(a,q)=1}}{}$, $\left[~\right]$ is the Iverson bracket and $e(z)=e^{2\pi iz}$ as usual.

Numerical Approximations: Set $f_m(z):=\sum_{n=1}^{\infty}{\sigma_m(n)\,e(nz)},~~~$ $g_m(z):=\frac{B_{m+1}}{2(m+1)}\left(1-\frac{1}{z^{m+1}}\right)+\left[m=1\right]\frac{1}{2}\frac{1}{2\pi iz}$

\begin{equation*} \begin{aligned} f_1\left(\frac{i}{2}\right)&=0.04916444073\\ g_1\left(\frac{i}{2}\right)&=0.04917839024\\~\\ f_3\left(\frac{i}{3}\right)&=0.3333338608\\ g_3\left(\frac{i}{3}\right)&=0.3333333333\\~\\ f_5\left(\frac{i}{3}-\frac{1}{10}\right)&=-0.1956996617 - 1.099266745*I\\ g_5\left(\frac{i}{3}-\frac{1}{10}\right)&=-0.1957162803 - 1.099271842*I\\~\\ f_7\left(\frac{i}{\sqrt{7}}+\frac{1}{5}\right)&=-1.365047865 - 1.273191433*I\\ g_7\left(\frac{i}{\sqrt{7}}+\frac{1}{5}\right)&=-1.363026857 - 1.272859570*I\\~\\ \end{aligned} \end{equation*}

Another surprising fact is that $g_1(z),\,g_3(z)$ and $g_5(z)$ satisfy Ramanujan's differential equations for $L(q),\,M(q),\,N(q)$ (as usual $q=e(z)$) as well!

By the way, the corresponding result for Eisenstein series is

\begin{equation} \boxed{\sum_{\substack{n,k\in\mathbb{Z}\\\left(n,k\right)\neq\left(0,0\right)}}^{\infty}{\frac{1}{\left(nz+k\right)^{m+1}}}\approx-\left(\frac{2\pi i}{z}\right)^{m+1}\frac{B_{m+1}}{(m+1)!}+\left[m=1\right]\frac{1}{2z}} \end{equation}

Best, M.