Asymptotic expansion of Fourier coefficients of reciprocals of Eisenstein series

Abstract

In this paper we give a classification of the asymptotic expansion of the q-expansion of reciprocals of Eisenstein series $E_k$ of weight k for the modular group $\mathop {\mathrm{SL}}_2(\mathbb {Z})$. For $k \ge 12$ even, this extends results of Hardy and Ramanujan, and Berndt, Bialek, and Yee, utilizing the Circle Method on the one hand, and results of Petersson, and Bringmann and Kane, developing a theory of meromorphic Poincaré series on the other. We follow a uniform approach, based on the zeros of the Eisenstein series with the largest imaginary part. These special zeros provide information on the singularities of the Fourier expansion of $1/E_k(z)$ with respect to $q = e^{2 \pi i z}$.

1 Introduction

In this paper we provide a new approach to determine the main asymptotic growth terms in the Fourier expansion of the reciprocals $1/E_k$ of Eisenstein series of weight k:

$$\begin{aligned} \frac{1}{E_k(z)} = \sum _{n=0}^{\infty } \beta _k(n) \, q^n \quad (q:= e^{2 \pi i z}). \end{aligned}$$

We refer to [4], Chapter 15 for a very good introduction into the topic. Eisenstein series are defined by

$$\begin{aligned} E_k(z):= 1 - \frac{2k}{B_k} \sum _{n=1}^{\infty } \sigma _{k-1}\left( n\right) \, q^n. \end{aligned}$$

They are modular forms [15] on the upper half of the complex plane $\mathbb {H}$. The algebra of modular forms with respect to the modular group $\mathop {\mathrm{SL}}_2(\mathbb {Z})$ is generated by $E_4$ and $E_6$. As usual $B_k$ denotes the kth Bernoulli number and $\sigma _{\ell }\left( n\right) := \sum _{d \mid n} d^{\ell }$.

Hardy and Ramanujan [8] launched, in their last joint paper, the study of coefficients of meromorphic modular forms with a simple pole in the standard fundamental domain $\mathbb {F}$. They demonstrated that, similar to their famous asymptotic formula for the partition numbers

$$\begin{aligned} p(n) \sim \frac{1}{4n \sqrt{3}} \, e^{ \pi \sqrt{\frac{2}{3}n}}, \qquad \sum _{n=0}^{\infty } p(n) \, q^n := \frac{q^{\frac{1}{24}}}{\eta (z)}, \end{aligned}$$

which had been given birth to the Circle Method [7], formulas for the coefficients of reciprocals of modular forms can be obtained. The reciprocal of the Dedekind $\eta$-function is a weakly modular form of weight $-1/2$ on $\mathbb {H}$.

Hardy and Ramanujan focused on the reciprocal of the Eisenstein series $E_6$. They proved an explicit formula for the coefficients. Shortly afterwards, in a letter to Hardy, Ramanujan stated several formulas of the same type, including the q-expansion of $1/E_4$. No proofs were given.

Bialek in his Ph.D. thesis, written under the guidance of Berndt [2], and finally Berndt, Bialek, and Yee [3] have proven the claims in the letter of Ramanujan by extending the methods applied in [8].

We illustrate the case $k=4$. Following Ramanujan, we frequently put $\textit{E}_k(q_z) := E_k(z)$ for $q=q_{z} := e^{2 \pi i z}$. Let $\rho$ be the unique zero of $E_4$ in $\mathbb {F}$. Let $\lambda$ run over the integers of the form $3^{\alpha } \prod _{\ell =1}^{r} p_{\ell }^{\alpha _{\ell }}$, where $\alpha =0$ or 1. Here, $p_{\ell }$ is a prime of the form $6m+1$, and $\alpha _j \in \mathbb {N}_0$. Then [2]:

$$\begin{aligned} \beta _4(n)= (-1)^n \frac{3}{\textit{E}_6(q_{\rho })} \sum _{( \lambda )} \sum _{ (c,d)} \frac{h_{(c,d)}(n)}{\lambda ^3} \,\, e^{\frac{\pi n \sqrt{3}}{\lambda }}. \end{aligned}$$

(1)

Here, $(c,d)\ne (0,0)$ and coprime, runs over distinct solutions to $\lambda = c^2 - cd + d^2$. Let (a, b) be such that $ad-bc=1$. Let $h_{(1,0)}(n):=1$, $h_{(2,1)}(n):=(-1)^n$, and for $\lambda \ge 7$:

$$\begin{aligned} h_{(c,d)}(n):= 2 \mathop {\mathrm{cos}} \left( (ad+bc - 2 ac - 2bd + \lambda ) \frac{ \pi \, n}{\lambda } - 6 \mathop {\mathrm{arctan}} \left( \frac{c \sqrt{3}}{2d - c}\right) \right) . \end{aligned}$$

For the definition of distinct we refer to [2,  Sect. 3]. From the explicit formula (1) one observes that the main asymptotic growth comes from $(c,d)=(1,0)$. This yields ([5], Introduction):

$$\begin{aligned} \beta _{4}(n)\sim & {} (-1)^n \frac{3}{E_6(\rho )} \, e^{\pi n \sqrt{3}}, \end{aligned}$$

(2)

$$\begin{aligned} \beta _{6}(n)\sim & {} \frac{2}{E_8(i)} \, e^{2 \pi n } , \end{aligned}$$

(3)

where $\sum _{n=0}^{\infty } \beta _k(n) \, q^n := \frac{1}{E_{k}\left( z\right) }$. We added the asymptotic (3), which can be obtained in a similar way.

Table 1 Quotients of successive coefficients of $1/\textit{E}_{k}$ for $k\in \left\{ 4,6,12,14\right\}$

Petersson [16] offered an alternative approach to study the q-expansion of meromorphic modular forms. He defined Poincaré series with poles at arbitrary points in $\mathbb {H}$ and of arbitrary order, to provide a basis for the underlying vector spaces. Recently, Bringmann and Kane [5] have generalized Petersson’s method. They have also recorded several important examples. In this paper we study the asymptotic expansions for all reciprocals of Eisenstein series. Instead of proving first an explicit formula and then detecting the main growth terms, we provide a direct approach. This is based on the distribution of the zeros in the standard fundamental domain with the largest imaginary part. For the convenience of the reader, we recall some basic idea from complex analysis ([20, 17,  Chap. 7, Sect. 5, task 242]). Let $f(q)=\sum _{n=0}^{\infty } a(n) \, q^n$ be a power series regular at $q=0$ with finite radius of convergence. Assume that there is only one singular point $q_0$ on the circle of convergence. Let at $q_0$ be a pole. Then it is known that

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{a(n)}{a(n+1)} = q_0. \end{aligned}$$

(4)

This follows from the Laurent expansion of f(q), which has a finite principal part.

Before we state our results, we want to point out as a warning that the limits as $n \rightarrow \infty$ for $\beta _{4}\left( n\right) / \beta _{4}\left( n+1\right)$ and $\beta _{6}\left( n\right) / \beta _{6}\left( n+1\right)$ exist, but that this is maybe not true for all k as provided by the data in Table 1.

2 Results

The constants in the asymptotic expansion of $\beta _{k}(n)$, the coefficients of the q-expansion of the reciprocal of $\textit{E}_k$, involve the Ramanujan $\Theta$-operator [6, 18] induced by residue calculation. The differential operator $\Theta := q \frac{\mathrm {d}}{\mathrm {d}q}$ acts on formal power series by

$$\begin{aligned} \Theta \left( \sum _{n=h}^{\infty } a(n) \, q^n \right) := \sum _{n=h}^{\infty } n \, a(n) \, q^n. \end{aligned}$$

Let $\textit{E}_2(q):= 1 -24 \sum _{n=1}^{\infty } \sigma _1(n) \, q^n$. Ramanujan observed that

$$\begin{aligned} \Theta (\textit{E}_4) = \left( \textit{E}_4 \textit{E}_2 - \textit{E}_6 \right) /3 \text { and } \Theta (\textit{E}_6) = \left( \textit{E}_6 \textit{E}_2 - \textit{E}_8 \right) /2. \end{aligned}$$

Our first results give an explicit interpretation of the data presented in Table 1 for $k=6$ and $k=14$.

Theorem 1

Let $k \ge 4$ and $k \equiv 2 \pmod {4}$ be an integer. Then $1/\textit{E}_k$ has a q-expansion with radius $q_{i}=e^{-2 \pi }$:

$$\begin{aligned} \frac{1}{ \textit{E}_k(q)} = \sum _{n=0}^{\infty } \beta _k(n) \, q^n. \end{aligned}$$

The coefficients $\beta _k(n)$ are non-zero and have the asymptotic expansion

$$\begin{aligned} \beta _k(n) \sim - \frac{1}{ \Theta (\textit{E}_k)(q_{i})}\,\,q_i^{-n}. \end{aligned}$$

The number $q_i= e^{-2 \pi } \approx 1.867443 \cdot 10^{-3}$ is transcendental. It is well-known that the so-called Gel$^{\prime }$fond constant $e^{\pi }$ is transcendental. This was first proven by Gel$^{\prime }$fond in 1929. It can also be deduced from the Gel$^{\prime }$fond–Schneider Theorem, which solved Hilbert’s seventh problem [21]. We refer to a result by Nesterenko (also [21,  Sect. 5.6]). Let $z \in \mathbb {H}$. Then at least already three of the four numbers

$$\begin{aligned} q_z, \textit{E}_2(q_z), \textit{E}_4(q_z), \text { and } \textit{E}_6(q_z) \end{aligned}$$

are algebraically independent. Since $\textit{E}_4(q_{\rho }) = \textit{E}_6(q_i)=0$, we obtain that $q_i, \textit{E}_4(q_i)$ and $q_{\rho }, \textit{E}_6(q_{\rho })$ are transcendental.

Moreover, $\Theta (\textit{E}_k)(q_{i})$ for $k=6,10,14$ can be explicitly expressed by $\Gamma (\frac{1}{4})$ and $\pi$. For example,

$$\begin{aligned} \Theta (\textit{E}_6)(q_{i}) = - \frac{1}{2} \textit{E}_4(q_i)^2,\quad \text { where } \textit{E}_4(q_i)= \frac{3 \, \Gamma (\frac{1}{4})^8}{(2 \pi )^6}. \end{aligned}$$

We can also extract the numbers $q_i$ and $\textit{E}_4(q_i)$ from the coefficients.

Corollary 1

Let $k \ge 4$ and $k \equiv 2 \pmod {4}$. Then

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{\beta _{k}\left( n\right) }{\beta _{k}\left( n+1\right) }= & {} q_{i}, \end{aligned}$$

(5)

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{\beta _{6}\left( n\right) }{\beta _{10}\left( n\right) }= & {} \lim _{n \rightarrow \infty } \frac{\beta _{10}\left( n\right) }{\beta _{14}\left( n\right) } = \textit{E}_4(q_i). \end{aligned}$$

(6)

Hardy and Ramanujan stated lower and upper bounds at the end of their initial work [8] on the coefficients of the reciprocal of $1/E_6$. We generalize their idea to all cases $k \equiv 2 \pmod {4}$ including $k=2$ and also improve their result in the original case $k=6$.

Theorem 2

Let $k \equiv 2 \pmod {4}$ and k a positive integer. Let $x(k):= \frac{2k}{B_k}$. Then we have for all $n \in \mathbb {N}$

$$\begin{aligned} \frac{\left( \frac{x(k)+\sqrt{\Delta _{k}}}{2} \right) ^{n+1}-\left( \frac{x(k)-\sqrt{\Delta _{k}}}{2} \right) ^{n+1} }{\sqrt{\Delta _{k}}} \le \beta _k(n) \end{aligned}$$

with $\Delta _{k}=x(k)^{2}+4\left( 2^{k-1}+1\right) x\left( k\right)$ and

$$\begin{aligned} \beta _k(n) \le \frac{\left( x(k)-\frac{b_{k}-\sqrt{D_{k}}}{2}\right) \left( \frac{b_{k}+\sqrt{D_{k}}}{2}\right) ^{n}+\left( \frac{b_{k}+\sqrt{D_{k}}}{2}-x(k)\right) \left( \frac{b_{k}-\sqrt{D_{k}}}{2}\right) ^{n}}{\sqrt{D_{k}}} \end{aligned}$$

(7)

with $b_{k}=x\left( k\right) +a_{k}$, $c_{k}=\left( 2^{k-1}+1-a_{k}\right) x\left( k\right)$, and $D_{k}=b_{k}^{2}+4c_{k}$ for all k where $a_{2}=\sqrt{7/3}$ and $a_{k}=\frac{3^{k-1}+1}{2^{k-1}+1}$ for $k\ge 6$.

The case $k\equiv 0 \pmod {4}$ is more complicated. For large k, we cannot expect that the limit as ${n \rightarrow \infty }$ of $\beta _k(n)/\beta _{k} (n+1)$ exists, since we have two poles on the circle of convergence. But for $k=4$ and $k=8$ there is still only one pole.

Proposition 1

Let $q_{\rho } = e^{2 \pi \rho } = - e^{ - \pi \sqrt{3}}$. Let $m \in \mathbb {N}$. Then the coefficients $\beta _{4,m}(n)$ of the mth power of $\textit{E}_4^{-1}$ i.e.

$$\begin{aligned} \sum _{n=0}^{\infty } \beta _{4,m}(n) \, q^n := \left( \frac{1}{\textit{E}_4(q)}\right) ^m \end{aligned}$$

satisfy for all m:

$$\begin{aligned} \lim _{ n \rightarrow \infty } \frac{\beta _{4,m}(n)}{\beta _{4,m}(n+1)} = q_{\rho }. \end{aligned}$$

Remarks

(a) For small weights the following identities exist:

$$\begin{aligned} E_8 = E_4^2, \, E_{10}= E_4 \cdot E_6 \text { and }E_{14} = E_4^2 \cdot E_6. \end{aligned}$$

(8)

(b) Let the principal part of $\textit{E}_4^{-m}$ at the pole $q_{\rho }$ be given by

$$\begin{aligned} \sum _{k=1}^m \frac{\lambda _{m,k}}{\left( q - q_{\rho }\right) ^{k}} , \end{aligned}$$

(9)

then $\lambda _{m,m} = {\mathop {\mathrm{res}}}_{q_{\rho }}\left( \textit{E}_4^{-1}\right) ^m.$ It would be interesting to get explicit formulas for all $\lambda _{m,k}$, $1 \le k \le m$. Especially for the case $m=2$. (c) We have $\mathop {\mathrm{res}}_{q_{\rho }} (\textit{E}_4^{-1}) = \frac{-3\, q_{\rho }}{\textit{E}_6(q_{\rho })}$.

We know that $\beta _4(n)$ and $\beta _8(n)$ are non-zero for all $n \in \mathbb {N}_0$ [9]. We provide new proof of the asymptotic expansion for $k=4$. This is the main term of a formula first conjectured by Ramanujan and proven about 80 years later by Bialek [2]. For the case $k=8$, we also refer to [5].

Theorem 3

We have $(-1)^n \beta _4(n) \in 240 \mathbb {N}$ for all $n \in \mathbb {N}$. Further, we have the asymptotic expansion

$$\begin{aligned} \beta _4(n) \sim - \frac{1}{\Theta (\textit{E}_4) ({q_{\rho }})} \, q_{\rho }^{-n}, \end{aligned}$$

where $\Theta (\textit{E}_4) ({q_{\rho }}) = -\textit{E}_6(q_{\rho })/3$.

F. K. C. Rankin and H. P. F. Swinnerton-Dyer [19] have proven that all the zeros of $E_k(z)$ in the standard fundamental domain $\mathbb {F}$ are in $C=\{ z \in \mathbb {F}\, : \, \vert z \vert =1 \} \subset \mathbb {F}$. We recall the following basic facts [15]. The modular group $\Gamma :=\mathop {\mathrm{SL}}_2(\mathbb {Z})$ operates on the complex upper half plane $\mathbb {H}$, denoted by $\gamma (z)$, where $\gamma \in \Gamma$ and $z \in \mathbb {H}$. The standard fundamental domain $\mathbb {F}$ is given by

$$\begin{aligned}\mathbb {F}= & {} \left\{ z \in \mathbb {H}\, : \, \vert z \vert \ge 1 \text { and } 0 \le \mathop {\mathrm{Re}}\left( z\right) \le 1/2 \right\} \\&\quad \cup \left\{ z \in \mathbb {H}\, : \, \vert z \vert > 1 \text { and } - 1/2< \mathop {\mathrm{Re}}\left( z\right) < 0 \right\} . \end{aligned}$$

Proposition 2

(Rankin, Swinnerton-Dyer [19]) Let $k \ge 4$ be an even integer. Let $z_k$ be the zero of $E_k$ with the largest imaginary part. Then

$$\begin{aligned} z_4 = z_8 = \rho ~{\mathrm{and} } ~z_k = i~\mathrm{for }~k \equiv 2 \pmod {4}. \end{aligned}$$

All other k satisfy $z_k \in C\, \backslash \, \{ i, \rho \}$. Only for $k=8$ the zero $z_k$ is not simple.

Further, from [19] and Kohnen [12] we obtain the following.

Corollary 2

Let $k \ge 12$ and $k \equiv 0 \pmod {4}$. Let $k = 12 \, N + s$ for $s \in \{ 0,4,8\}$. Then $z_k = e^{\frac{1}{2}\pi i \, \varphi }$, where $\varphi \in \left( \frac{N-1}{N},1 \right)$.

Theorem 4

Let k be a positive integer. Let $k \ge 12$ and $k \equiv 0 \pmod {4}$. Then $1/\textit{E}_k$ has a q-expansion with radius $\vert q_{z_k} \vert$, where $z_k$ is the zero of $E_k$ with the largest imaginary part. Then

$$\begin{aligned} \beta _k(n) \, q_{z_k}^{n} + \frac{1}{ \Theta (\textit{E}_k)(q_{z_k})} + \frac{1}{ \Theta (\textit{E}_k)(\overline{q}_{z_k})} \left( \frac{q_{z_k}}{\overline{q}_{z_k}}\right) ^{n} \end{aligned}$$

constitutes a null sequence.

The expression

$$\begin{aligned} \frac{1}{ \Theta (\textit{E}_k)(q_{z_k})} + \frac{1}{ \Theta (\textit{E}_k)(\overline{q}_{z_k})} \left( \frac{q_{z_k}}{\overline{q}_{z_k}}\right) ^{n} \end{aligned}$$

(10)

is bounded. But this is not sufficient to obtain an asymptotic expansion. Nevertheless we have discovered a new property of the coefficients of $1/ \textit{E}_k$ for $k \equiv 0 \pmod {4}$.

Theorem 5

Let $k \equiv 0 \pmod {4}$ and $k \ge 12$. Then there exists a subsequence $\{n_t\}_{t=1}^{\infty }$ of $\{n\}_{n=1}^{\infty }$ such that

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{\beta _k(n_t)}{ -q_{z_k}^{-n_t} \left( \frac{1}{ \Theta (\textit{E}_k)(q_{z_k})} + \frac{1}{ \Theta (\textit{E}_k)(\overline{q}_{z_k})} \left( \frac{q_{z_k}}{\overline{q}_{z_k}}\right) ^{n_t} \right) } = 1. \end{aligned}$$

The statement of this theorem is equivalent to

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{\beta _k(n_t) }{- 2 \mathop {\mathrm{Re}} \left( \frac{q_{z_k}^{-n_t}}{ \Theta (\textit{E}_k)(q_{z_k})}\right) } = 1. \end{aligned}$$

We have the following further properties.

Theorem 6

Let k be a positive integer. Let $k \ge 12$ and $k \equiv 0 \pmod {4}$.

• (1) Let $A_k(n)$ denote the number of changes of sign in the sequence $\{\beta _k(m)\}_{m=0}^n$ and let $z_k = x_k + i \, y_k \in \mathbb {F}$ be the zero of $E_k$ with the largest imaginary part. Then

  $$\begin{aligned} \lim _{ n \rightarrow \infty } \frac{A_k(n)}{n} = 2 x_k. \end{aligned}$$

• (2) Let $B_k(n)$ be the number of non-zero coefficients among the n coefficients $\{\beta _k(m)\}_{m=0}^{n-1}$. Then

  $$\begin{aligned} \limsup _{ n \rightarrow \infty } \frac{n}{B_k(n)} \le 2. \end{aligned}$$

Combining Theorem 6 (1) with Corollary 2 leads to the following result, which is a priori surprising.

Corollary 3

For large weights k divisible by 4, the coefficients of $1/ \textit{E}_k(q)$ satisfy

$$\begin{aligned} \lim _{\ell \rightarrow \infty } \lim _{n \rightarrow \infty } \frac{A_{4\ell }(n)}{n} =0. \end{aligned}$$

3 Proofs

3.1 Proof of Proposition 1, Theorem 1, and Corollary 1

Let $\textit{E}_k(q)$ have exactly one zero $q_0 \in B_1(0)$ with absolute value smaller than all other zeros. Then we obtain the property (4) for the coefficients of $1/\textit{E}_k$. Note that every zero of a modular form has one representative in the fundamental domain $\mathbb {F}$.

The zeros of $E_k$ are controlled by a theorem by Rankin and Swinnerton-Dyer ([19], see also Sect. 2). They proved that every zero in $\mathbb {F}$ has absolute value 1. Further, let k be a positive, even integer and $k \ge 4$. Let $k = 12 N +s$, where $s \in \{ 4,6,8,10,0, 14\}$. Then $E_k$ has N simple zeros in $C\setminus \{i, \rho \}$. Additionally we have simple zeros $\rho$ for $s=4$ and i for $s=6$. Further, $E_k$ has the double zero $\rho$ for $s=8$, the simple zeros i and $\rho$ for $s=10$, and the simple zero i and the double zero $\rho$ for $s=14$. Further, let $z_k$ be the zero of $E_k$ with the largest imaginary part. Note that

$$\begin{aligned} z_k' := S \left( z_k\right) = \left( \begin{array}{cc} 0 &{} -1 \\ 1 &{} 0 \end{array}\right) z_k \end{aligned}$$

and $z_k$ have the same imaginary part. Note that $S\left( i\right) = i$ and $S\left( \rho \right) = \rho -1$. Thus, $1/\textit{E}_k$ has exactly one pole on the radius of convergence iff $z_k= i$ or $z_k = \rho$.

Proof of Proposition 1

Since $\left( 1/\textit{E}_{4}\right) ^m$ has only the pole at $q_{\rho }$ on the circle of convergence, again we have formula (4), which proves the proposition. $\square$

Proof of Theorem 1

Let w be any complex number. Let $B_r(w)= \{ z \in \mathbb {C} \, : \, \vert z - w \vert < r\}$ be the open ball with radius r around w. We denote the closure by $\overline{B_{r}\left( w\right) }$ and its boundary by $\partial B_{r}\left( w\right)$. Let $k \equiv 2 \pmod {4}$. Then $\textit{E}_k$ has the special property that restricted to $\overline{B_{\vert q_i \vert }(0)}$ it has exactly one zero at $q_i$, which is also simple. This implies that the Taylor series expansion of the reciprocal of $\textit{E}_k$ has radius of convergence $\left| q_{i}\right|$ and only a simple pole at $q_i$:

$$\begin{aligned} \frac{1}{ \textit{E}_k(q)} = \sum _{n=0}^{\infty } \beta _k(n) \, q^n \qquad ( \vert q \vert < \vert q_i \vert ). \end{aligned}$$

Note that subtracting the principal part at $q_i$ provides a new Taylor series expansion with a larger radius of convergence:

$$\begin{aligned} \frac{1}{ \textit{E}_k(q)} - \frac{ \mathop {\mathrm{res}}_{q_i} (1/\textit{E}_k)}{q - q_i} = \sum _{n=0}^{\infty } b(n) \, q^n. \end{aligned}$$

This implies that $b(n) q_i^n$ constitutes a null sequence. Here, $\mathop {\mathrm{res}}_{q_i} (1/\textit{E}_k)$ denotes the residue at $q_i$. We obtain that

$$\begin{aligned} q_i^{n+1} \beta _k(n) + \mathop {\mathrm{res}}{}_{q_i} (1/\textit{E}_k) \end{aligned}$$

constitutes a null sequence. By a standard argument, we obtain that

$$\begin{aligned} \mathop {\mathrm{res}}{}_{q_i} (1/\textit{E}_k) = \frac{1}{\frac{\mathrm {d}}{\mathrm {d}q} \textit{E}_k (q_i)}. \end{aligned}$$

Finally, we obtain the asymptotic behavior

$$\begin{aligned} \beta _k(n) \sim - \frac{1}{\Theta (\textit{E}_k)(q_i)}\, q_i^{-n}. \end{aligned}$$

$\square$

Table 2 Quotients of successive coefficients of $1/\textit{E}_{k}$ for $k\in \left\{ 8,10,12,14,16\right\}$

Table 3 Quotients of $\beta _{6}\left( n\right)$ and $\beta _{10}\left( n\right)$

Proof of Corollary 1

From the theorem of Rankin and Swinnerton-Dyer we obtain that for $k \equiv 2 \pmod {4}$ we have $z_k=i$ and $q_i = e^{-2 \pi }$. This gives a first proof of equation (5) of Corollary 1. Note that equation (5) of Corollary 1 also follows directly from Theorem 1. The quotients for small k converge very quickly. We refer to Table 1 and Table 2.

Since $\Theta ( \textit{E}_6) (q_i) = - \frac{1}{2} \textit{E}_4(q_i)^2$ and $\Theta ( \textit{E}_{10}) (q_i) = - \frac{1}{2} \textit{E}_4(q_i)^3$, the second part of the Corollary also follows from Theorem 1 and (8). An approximate numerical value of $\textit{E}_4(q_i)$ can be read off Table 3. The theorem by Nesterenko implies that this number is transcendental, since $\textit{E}_6(q_i)=0$. $\square$

Note that for each integer $\ell \ge 2$, the limit as $n \rightarrow \infty$ of $\frac{\beta _{4\ell -2}(n)}{\beta _{4\ell +2}(n)}$ exists, but it is generally not equal to $\textit{E}_4(q_i)$.

3.2 Proof of Theorem 2

We use the following easy to prove lemmata.

Lemma 1

$\sigma _{\ell }\left( n\right) <\frac{\ell }{\ell -1}n^{\ell }$ for $\ell >1$ and $\sigma _{1}\left( n\right) \le \left( 1+\ln n\right) n$.

Proof

$\sigma _{\ell }\left( n\right) \le \left( 1+\int _{1}^{n}t^{-\ell }\,\mathrm {d}t\right) n^{\ell }<\frac{\ell }{\ell -1}n^{\ell }$ for $\ell >1$ and $\le \left( 1+\ln n\right) n$ for $\ell =1$.    $\square$

Lemma 2

For $\ell \ge 5$ it holds that $3\root \ell \of {\frac{1+3^{-\ell }}{1+2^{-\ell }}}>2.98$.

Proof

Considering $\ell$ as a real variable $\ge 5$, we obtain the following logarithmic derivative

$$\begin{aligned}&\frac{\mathrm {d}}{\mathrm {d}\ell }\frac{1}{\ell }\ln \left( \frac{1+3^{-\ell }}{1+2^{-\ell }}\right) \\&\quad =-\frac{1}{\ell ^{2}}\ln \left( \frac{1+3^{-\ell }}{1+2^{-\ell }}\right) +\frac{1}{\ell }\frac{1+2^{-\ell }}{1+3^{-\ell }}\left( -\frac{3^{-\ell }\ln 3}{1+3^{-\ell }}+\frac{2^{-\ell }\ln 2}{1+2^{-\ell }}\right) >0 \end{aligned}$$

since $-\frac{\ln 3}{3^{\ell }+1}+\frac{\ln 2}{2^{\ell }+1}>-\frac{\ln 3}{3^{\ell }}+\frac{\ln 2}{2^{\ell +1}}>0$ for $\ell \ge 5$. Therefore, the values of the original sequence are increasing and we take the smallest value for $\ell =5$. $\square$

Proof of Theorem 2

With $\varepsilon _{k}\left( n\right) =\frac{2k}{B_{k}}\sigma _{k-1}\left( n\right)$ we obtain

$$\begin{aligned} E_{k}\left( z\right) =1-\sum _{n=1}^{\infty }\varepsilon _{k}\left( n\right) q^{n}. \end{aligned}$$

Let $1/\left( 1-\varepsilon _{k}\left( 1\right) q-\varepsilon _{k}\left( 2\right) q^{2}\right) =\sum _{n=0}^{\infty }\alpha _{k}\left( n\right) q^{n}$. The $\alpha _{k}\left( n\right)$ fulfill the recurrence relation $\alpha _{k}\left( n\right) =\varepsilon _{k}\left( 1\right) \alpha _{k}\left( n-1\right) +\varepsilon _{k}\left( 2\right) \alpha _{k}\left( n-2\right)$ for $n\ge 2$. Obviously, $\alpha _{k}\left( 0\right) =\beta _{k}\left( 0\right)$, $\alpha _{k}\left( 1\right) =\beta _{k}\left( 1\right)$, and by induction $\alpha _{k}\left( n\right) =\varepsilon _{k}\left( 1\right) \alpha _{k}\left( n-1\right) +\varepsilon _{k}\left( 2\right) \alpha _{k}\left( n-2\right) \le \sum _{j=1}^{n}\varepsilon _{k}\left( j\right) \beta _{k}\left( n-j\right) =\beta _{k}\left( n\right)$ using the power series expansion of $1/\textit{E}_{k}$.

For the upper bound let $a_{2}=\sqrt{7/3}$ and for $k\ge 6$ let

$$\begin{aligned} a_{k}=\frac{\varepsilon _{k}\left( 3\right) }{\varepsilon _{k}\left( 2\right) }=\frac{\sigma _{k-1}\left( 3\right) }{\sigma _{k-1}\left( 2\right) }=\frac{3^{k-1}+1}{2^{k-1}+1}. \end{aligned}$$

For all $k\equiv 2\pmod {4}$ let $b_{k}=a_{k}+\varepsilon _{k}\left( 1\right)$, $c_{k}=-\varepsilon _{k}\left( 2\right) -a_{k}\varepsilon _{k}\left( 1\right)$, and $\frac{1-b_{k}q-c_{k}q^{2}}{1-a_{k}q}=1-\sum _{n=1}^{\infty }\delta _{k}\left( n\right) q^{n}$. Therefore, $\delta _{k}\left( 1\right) =b_{k}-a_{k}=\varepsilon _{k}\left( 1\right)$, $\delta _{k}\left( 2\right) =c_{k}+a_{k}\delta _{k}\left( 1\right) =\varepsilon _{k}\left( 2\right)$, and $\delta _{k}\left( n\right) =a_{k}\delta _{k}\left( n-1\right)$ for $n\ge 3$. Therefore $\delta _{k}\left( n\right) =\varepsilon _{k}\left( 2\right) a_{k}^{n-2}$.

• (1) First, let $k=2$. Then $\delta _{2}\left( n\right) =72\left( 7/3\right) ^{\left( n-2\right) /2}$. For $n\in \left\{ 3,4,5,6\right\}$ we obtain $24\sigma _{1}\left( n\right) \le \delta _{2}\left( n\right)$. Using Lemma 1 we obtain $\varepsilon _{2}\left( n\right) \le 24\left( 1+\ln n\right) n$. For $n=7$ we obtain $24\cdot \left( 1+\ln 7\right) \cdot 7<504<72\left( 7/3\right) ^{\left( 7-2\right) /2}$ and for $n\ge 7$ we obtain $\frac{1+\ln \left( n+1\right) }{1+\ln n}\frac{n+1}{n}\le \left( 1+\frac{\ln \left( 1+\frac{1}{7}\right) }{1+\ln n}\right) \frac{8}{7}<1.2<\sqrt{7/3}$. Therefore, $\varepsilon _{2}\left( n\right) \le \delta _{2}\left( n\right)$.

• (2) Now, for $k\ge 6$

  $$\begin{aligned} \delta _{k}\left( n\right) =\varepsilon _{k}\left( 3\right) a_{k}^{n-3}=\frac{2k}{B_{k}}\left( 3^{k-1}+1\right) \left( \left( \frac{3}{2}\right) ^{k-1}\frac{1+3^{1-k}}{1+2^{1-k}}\right) ^{n-3}. \end{aligned}$$

  Using Lemma 1 we obtain $\sigma _{k-1}\left( n\right) <\frac{k-1}{k-2}n^{k-1}$. Since $k\ge 6$, Bernoulli’s inequality implies that $\frac{k-1}{k-2}\le \frac{5}{4}=1+\frac{1}{4}<\left( 1+\frac{1}{20}\right) ^{5} \le \left( \frac{21}{20}\right) ^{k-1}$. Therefore

  $$\begin{aligned} \root k-1 \of {\frac{B_{k}}{2k}\varepsilon _{k}\left( n\right) }=\root k-1 \of {\sigma _{k-1}\left( n\right) }<\root k-1 \of {\frac{k-1}{k-2}n^{k-1}}<\frac{21}{20}n. \end{aligned}$$

  Using Lemma 2 implies $\root k-1 \of {\frac{B_{k}}{2k}\delta _{k}\left( n\right) }>2.98\left( \frac{3}{2}\right) ^{n-3}$. Now $\frac{21}{20}n<2.98\left( \frac{3}{2}\right) ^{n-3}$ for $n\ge 4$ as $4.2<4.47$ for $n=4$ and $\frac{n}{n-1}<\frac{3}{2}$ for $n>4$.

We have shown $\varepsilon _{k}\left( n\right) =\delta _{k}\left( n\right)$ for $n\in \left\{ 1,2\right\}$ and $\varepsilon _{k}\left( n\right) \le \delta _{k}\left( n\right)$ for all $n\ge 3$. Let now $\frac{1-a_{k}q}{1-b_{k}q-c_{k}q^{2}}=\sum _{n=0}^{\infty }\gamma _{k}\left( n\right) q^{n}$. Then $\beta _{k}\left( n\right) =\gamma _{k}\left( n\right)$ for $n\in \left\{ 1,2\right\}$ and by induction $\gamma _{k}\left( n\right) =\sum _{j=1}^{n}\delta _{k}\left( j\right) \gamma _{k}\left( n-j\right) \ge \sum _{j=1}^{n}\varepsilon _{k}\left( j\right) \beta _{k}\left( n-j\right) =\beta _{k}\left( n\right)$ for $n\ge 3$.

We have shown $\alpha _{k}\left( n\right) \le \beta _{k}\left( n\right) \le \gamma _{k}\left( n\right)$ for all $n\ge 1$. From the generating functions we can now determine formulas for $\alpha _{k}\left( n\right)$ and $\gamma _{k}\left( n\right)$. The characteristic equation for $\alpha _{k}\left( n\right)$ is $\lambda _{k}^{2}-\varepsilon _{k}\left( 1\right) \lambda _{k}-\varepsilon _{k}\left( 2\right) =0$. Let $\Delta _{k}=\varepsilon _{k}\left( 1\right) ^{2}+4\varepsilon _{k}\left( 2\right) =\left( \frac{2k}{B_{k}}\right) ^{2}+\frac{8k}{B_{k}}\left( 2^{k-1}+1\right)$. Then $\lambda _{k,\pm }=\frac{1}{2}\left( \varepsilon _{k}\left( 1\right) \pm \sqrt{\Delta _{k}}\right)$. We obtain $\left( \begin{array}{c} L_{k,+} \\ L_{k,-} \end{array} \right) =\left( \begin{array}{cc} 1 &{} 1 \\ \lambda _{k,+} &{} \lambda _{k,-} \end{array} \right) ^{-1}\left( \begin{array}{c} 1 \\ \varepsilon _{k}\left( 1\right) \end{array} \right) =\frac{1}{\lambda _{k,+}-\lambda _{k,-}}\left( \begin{array}{c} \varepsilon _{k}\left( 1\right) -\lambda _{k,-} \\ \lambda _{k,+}-\varepsilon _{k}\left( 1\right) \end{array} \right) =\frac{1}{\sqrt{\Delta _{k}}}\left( \begin{array}{c} \lambda _{k,+} \\ -\lambda _{k,-} \end{array} \right)$. Therefore, $\alpha _{k}\left( n\right) =L_{k,+}\lambda _{k,+}^{n}+L_{k,-}\lambda _{k,-}^{n}=\frac{\lambda _{k,+}^{n+1}-\lambda _{k,-}^{n+1}}{\sqrt{\Delta _{k}}}$ for all n.

The characteristic equation for $\gamma _{k}\left( n\right)$ is $\mu _{k}^{2}-b_{k}\mu _{k}-c_{k}=0$. Let $D _{k}=b_{k}^{2}+4c_{k}$. Then $\mu _{k,\pm }=\frac{1}{2}\left( b_{k}\pm \sqrt{D_{k} }\right)$,

$$\begin{aligned} \left( \begin{array}{c} M_{k,+} \\ M_{k,-} \end{array} \right) =\left( \begin{array}{cc} 1 &{} 1 \\ \mu _{k,+} &{} \mu _{k,-} \end{array} \right) ^{-1}\left( \begin{array}{c} 1 \\ \varepsilon _{k}\left( 1\right) \end{array} \right) =\frac{1}{\sqrt{D_{k} }} \left( \begin{array}{c} \varepsilon _{k}\left( 1\right) -\mu _{k,-} \\ \mu _{k,+}-\varepsilon _{k}\left( 1\right) \end{array} \right) , \end{aligned}$$

and $\gamma _{k}\left( n\right) =M_{k,+}\mu _{k,+}^{n}+M_{k,-}\mu _{k,-}^{n}$. $\square$

Example

(Slight improvement of [8]) Let $k=6$. Then

$$\begin{aligned} \alpha _{6}\left( n\right)= & {} \frac{1}{\sqrt{320544}}\left( \left( \frac{504+\sqrt{320544}}{2}\right) ^{n+1}-\left( \frac{504-\sqrt{320544}}{2}\right) ^{n+1}\right) \\\approx & {} \frac{1}{566.16}\left( 535.08^{n+1}-\left( -31.083\right) ^{n+1}\right) . \end{aligned}$$

With $x\left( 6\right) =\frac{12}{B_{6}}=504$, $a_{6}=\frac{244}{33}$, $b_{6}=\frac{16876}{33}$, $c_{6}=\frac{141960}{11}$, $D_{6}=\frac{341015536}{1089}$ and $\sqrt{D_{6}}\approx 559.59$ we obtain $\mu _{6,\pm } = \frac{b_{6}\pm \sqrt{D_{6}}}{2}$,

$$\begin{aligned} M_{6,+} = \frac{1}{\sqrt{D_{6}}}\left( x{(6)}-\frac{b_{6}-\sqrt{D_{6}}}{2}\right) ,\qquad M_{6,-}=\frac{1}{\sqrt{D_{6}}}\left( \frac{b_{6}+\sqrt{D_{6}}}{2}-x{(6)}\right) . \end{aligned}$$

By (7) this finally yields

$$\begin{aligned} \gamma _{6}\left( n\right) =M_{6,+}\mu _{6,+}^{n}+M_{6,-}\mu _{6,-}^{n} \approx \frac{528.10\cdot 535.49^{n}+31.494\cdot \left( -24.100\right) ^{n}}{559.59} . \end{aligned}$$

The second and last column in Table 4 are the lower and upper bounds from [8].

Table 4 Improvement of upper and lower bounds (approximation) for $\beta _6(n)$

3.3 Proof of Theorem 3

For the special case of $k=4$ we refer to a result of [9]. We have proven that $(-1)^n \beta _4(n) \in 240 \, \mathbb {N}$ for all $n \in \mathbb {N}$ (see also [1], last section, for an announcement of the result of strict sign changes). We are mainly interested in the implication $\beta _4(n) \ne 0$.

Proof of Theorem 3

Let $k=4$. Then $z_4= \rho$ and $S\left( z_{4}\right) = \rho -1$. This implies that $1/\textit{E}_4(q) = \sum _{n=0}^{\infty } \beta _4(n) \, q^n$ has $\vert q_{\rho } \vert$ as the radius of convergence. Further, the only singularity on the circle of convergence is given by the pole at $q_{\rho }$. Now we can proceed as in the proof of Theorem 1 and obtain the asymptotic expansion of $\beta _4(n)$. Here we use the fact that $\mathop {\mathrm{res}}_{q_{\rho }} \textit{E}_4^{-1}$ is equal to

$$\begin{aligned} \frac{q_{\rho }}{\Theta \left( \textit{E}_{4}\right) \left( q_{\rho }\right) }= \frac{ -3 \, q_{\rho }}{\textit{E}_6(q_{\rho })}. \end{aligned}$$

$\square$

3.4 Proof of Theorem 4 and Theorem 5

Proof of Theorem 4

Let $k \equiv 0 \pmod {4}$. We are interested in the zeros of $E_k$ which contribute to poles on the circle of convergence of the power series

$$\begin{aligned} \frac{1}{\textit{E}_k(q)} = \sum _{n=0}^{\infty } \beta _k(n) \, q^n. \end{aligned}$$

Let $k \ge 12$ then Proposition 2 and Corollary 2 imply that there are precisely two singularities on the boundary of the region of absolute convergence, provided by the two poles at $q_{z_k}$ and $\overline{q}_{z_k}$. This implies that the radius of convergence is equal to $\left| q_{z_k} \right|$. Here we also used the well-known fact, that the imaginary part of $\gamma (z)$, when $\gamma$ is in the modular group and z in the fundamental domain, does not increase. Next we consider the Laurent expansion of $1 / \textit{E}_k(q)$ around $q_{z_k}$. We subtract the principal part from $1 / \textit{E}_k(q)$ and obtain a holomorphic function at $q_{z_k}$. We iterate this procedure and consider the Laurent expansion around the other pole $\overline{q}_{z_k}$ and subtract again the principal part. Note that we have poles of order one. This implies that

$$\begin{aligned} \frac{1}{\textit{E}_{k}\left( q\right) } - \frac{ \mathop {\mathrm{res}}_{q_{z_k}} \textit{E}_k^{-1}}{q - q_{z_k}} - \frac{ \mathop {\mathrm{res}}_{\overline{q}_{z_k}} \, \, \textit{E}_k^{-1}}{q - \overline{q}_{z_k}} \end{aligned}$$

(11)

has a holomorphic expansion $\sum _{n=0}^{\infty } b(n) \, q^n$, with a radius of convergence larger than $\vert q_{z_k}\vert = \left| \overline{q}_{z_k}\right|$. This implies that $b(n) q_{z_k}^n$ and $b(n) \overline{q}_{z_k}^{n}$ constitute null sequences. The residue values can be expressed by $\Theta (\textit{E}_k)$ evaluated at the poles. This leads to an expression which allows in the final formula the number $q_{z_k}^{-n}$ to appear instead of $q_{z_k}^{-(n+1)}$. See also the proof of Theorem 1. By the identity principle b(n) is equal to

$$\begin{aligned} \beta _{k}(n) + \frac{1}{\Theta \left( \textit{E}_k\right) (q_{z_k})} \, q_{z_k}^{-n} + \frac{1}{\Theta \left( \textit{E}_k\right) (\overline{q}_{z_k})} \, \overline{q}_{z_k}^{-n}. \end{aligned}$$

This implies that

$$\begin{aligned} \sum _{n=0}^{\infty } \left( \beta _{k}(n) + \frac{1}{\Theta \left( \textit{E}_k\right) (q_{z_k})} \, q_{z_k}^{-n} + \frac{1}{\Theta \left( \textit{E}_k\right) (\overline{q}_{z_k})} \, \overline{q}_{z_k}^{-n} \right) q^{n} = \sum _{n=0}^{\infty } b(n) q_{z_k}^n \, \left( \frac{q}{{q}_{z_k}}\right) ^n \end{aligned}$$

for $q \in \mathbb {C}$ and $\left| q\right| < \vert q_{z_k} \vert$. Let $w= q/ q_{z_k}$. Then

$$\begin{aligned} \sum _{n=0}^{\infty } \left( \beta _{k}(n) q_{z_k}^{n} + \frac{1}{\Theta \left( \textit{E}_k\right) (q_{z_k})} + \frac{1}{\Theta \left( \textit{E}_k\right) (\overline{q}_{z_k})} \, \left( \frac{q_{z_k}}{ \overline{q}_{z_k}}\right) ^n \right) w^n = \sum _{n=0}^{\infty } b(n) q_{z_k}^n \, w^n . \end{aligned}$$

In the final step we compare the coefficients with respect to $w^{n}$ and use the identity principle for regular power series. Since $b(n) \, q_{z_k}^n$ constitutes a null sequence, the claim of the theorem follows. $\square$

Proof of Theorem 5

Let $k \equiv 0 \pmod {4}$ and $k \ge 12$. Let $z_k = x_k +iy_k$ be the zero of $E_k$ in $\mathbb {F}$ with the largest imaginary part. Then $z_k \ne i, \rho$. This implies by results by Kanou [10] and Kohnen [11] that $z_k$ is transcendental. Since we have chosen $z_k$ on the circle of unity, we can conclude that $x_k$ and $y_k$ are also transcendental. By a well-known result by Kronecker [14], since $x_k$ is irrational, the orbit

$$\begin{aligned} \mathbb {O}_k := \left\{ \left( \frac{ q_{z_k}}{\overline{q}_{z_k}}\right) ^n \, : n \in \mathbb {N} \right\} \end{aligned}$$

is dense in $\left\{ w= e^{2 \pi i \alpha } \, : \, \alpha \in [0,1) \right\}$. Let $C_k:= 1/ \Theta (\textit{E}_k)(q_{z_k})$. Since

$$\begin{aligned} \overline{C}_k= 1/ \Theta (\textit{E}_k)(\overline{q}_{z_k}), \end{aligned}$$

for the closure of the set

$$\begin{aligned} D_k:= \left\{ \frac{1}{ \Theta (\textit{E}_k)(q_{z_k})} + \frac{1}{ \Theta (\textit{E}_k)(\overline{q}_{z_k})} \left( \frac{q_{z_k}}{\overline{q}_{z_k}}\right) ^{n}\, : \, n \in \mathbb {N} \right\} , \end{aligned}$$

we obtain a circle with center $C_k$ and radius $\vert C_k\vert$:

We note that 0 and $2 \, C_k$ are not elements of $D_k$. Let $d_k \in \partial B_{\vert C_k \vert } (C_k)\setminus \left\{ 0\right\}$. Then there exists a subsequence $\left\{ n_{t}\right\} _{t=1}^{\infty }$ of $\{n\}_{n=1}^{\infty }$ such that

$$\begin{aligned} \lim _{ t \rightarrow \infty } \frac{1}{ \Theta (\textit{E}_k)(q_{z_k})} + \frac{1}{ \Theta (\textit{E}_k)(\overline{q}_{z_k})} \left( \frac{q_{z_k}}{\overline{q}_{z_k}}\right) ^{n_t} = d_k. \end{aligned}$$

Combining this result with Theorem 4 proves the claim. $\square$

3.5 Proof of Theorem 6 and Corollary 3

We recall a result from complex analysis. Pólya and Szegő recorded the following beautiful property ([17], Part Three, Chapter 5). Let $f(x)= \sum _{n=0}^{\infty } a(n) \, x^n$ be a power series with radius of convergence $0< r < \infty$ and real coefficients. We assume that we have only two singularities on the circle of convergence and that these two singularities are poles: $x_1 = r e^{i \alpha }$ and $x_2 = r e^{- i \alpha }$ with $0< \alpha < \pi$. Let A(n) denote the number of changes of sign in the sequence $\{a(m)\}_{m=0}^n$. Then $\lim _{n \rightarrow \infty } \frac{A(n)}{n} = \frac{\alpha }{\pi }$. The number of changes of sign in a sequence of real numbers is given by the sign changes of the sequence, when all zeros are removed. Results in this direction had also been given by König [13] in 1875.

Proof of Theorem 6, part (1)

Let $k \equiv 0 \pmod {4}$. Then $1/\textit{E}_k(q)= \sum _{n=0}^{\infty } \beta _k(n) \, q^n$ has a radius of convergence $\vert q_{z_k} \vert$, where $z_k = x_k + i y_k$ is the zero of $E_k$ with the largest imaginary part with $0< x_k < 1/2$. We stated already that $q_{z_k}$ and $\overline{q}_{z_k}$ are the single two singularities on the circle of convergence. Note that $q_{z_k} = r_k \cdot e^{2 \pi i x_k}$, where $r_k = e^{-2\pi y_k} = \vert q_{z_k} \vert$. Further, $\overline{q}_{z_k} = r_k \cdot e^{-2 \pi i x_k}$. Thus all assumptions are fulfilled to apply the above cited result for $A(n)= A_k(n)$ and $\alpha = 2 x_k$. $\square$

Example

We have $z_{16}\approx 0.196527+ 0.980498 \, i$. See Table 5 for values $A_{16}\left( n\right) /n$.

Table 5 Portion of sign changes for $k=16$

We also recall another interesting result stated in [17] (Part Three, Chap. 5). Let $f(x)= \sum _{n=0}^{\infty } a(n) \, x^n$ be a power series with finite positive radius of convergence. We assume that there are only poles on the circle of convergence. Let B(n) be the number of non-zero coefficients among the first n coefficients $\{a(m)\}_{m=0}^{n-1}$. Then the number of poles is not smaller than

$$\begin{aligned} \limsup _{ n \rightarrow \infty } \frac{n}{B(n)}. \end{aligned}$$

(12)

Proof of Theorem 6, part (2)

The number of poles is 2. Thus, by the result above, 2 is an upper bound for the term (12), which completes the proof. $\square$

Example

We have $B_{12}(n) = B_{16}(n) = B_{20}(n)=n$ for $n \le 1000$.

Proof of Corollary 3

From Theorem 6 we obtain

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{A_{4 \ell }}{4\ell } = 2 \, x_{4 \ell }, \end{aligned}$$

where $x_{4 \ell }$ is the real part of the zero of $E_{4 \ell }$ with the largest imaginary part. Finally, from Corollary 2 the claim follows, since $x_{4 \ell }$ tends to zero. $\square$