# Fourier series of Eisenstein series — elegant and very good approximation

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 $$0 near the imaginary axis - is still so good? (Even the correction term in case of $$E_2$$, i.e. for $$m=1$$, is matched!).

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

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

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

Best, M.

• If approach to the real axis stays sufficiently close to the imaginary axis, it is easily-directly related to "going to infinity" within the standard fundamental domain. In the latter case, either by general principles or by explicit estimates, the constant term strongly dominates the asymptotics... Is this the sort of thing that would be a useful answer to your question? Jan 25 at 22:04
• Hi Paul, Thanks for this - yes this is useful and answers my question. Jan 27 at 7:48

As Paul Garrett says, this reflects (for $$k \geq 4$$ even) the modularity of the Eisenstein series $$E_k(z) = -\frac{B_k}{2k} + \sum_{n=1}^{\infty} \sigma_{k-1}(n) e(nz)$$ with respect to the matrix $$S = (\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix})$$, which sends $$z$$ to $$-1/z$$ (so it preserves the imaginary axis and swaps $$0$$ and $$i\infty$$). For $$m \geq 3$$ odd, writing $$E_{m+1} |_{m+1} S = E_{m+1}$$ you get an asymptotic expansion of $$E_{m+1}(z)$$ at $$z=0$$, which agrees with what you wrote. For $$m=1$$ the Eisenstein series $$E_2$$ is only quasi-modular, but you can also work out the asymptotic expansion.
The integer $$m$$ should be odd, because $$E_k(z)$$ has no modularity property whatsoever for $$k$$ odd.
• Thanks, Francois. This is exactly what I was looking for. Also, if I interpret your answer right, this means that the power series expansion for the Bernoulli numbers is a valid step that delivers the right answer, but ONLY for z=0, which corresponds then to $i\infty$ under the transformation S. Jan 27 at 7:51