## 1 Introduction

As is well known, the Bernoulli polynomials $$B_{m}(x)$$ are given by the generating function

$$\frac{t}{e^{t}-1}e^{xt}=\sum_{m=0} ^{\infty}B_{m}(x)\frac{t^{m}}{m!}.$$

The Genocchi polynomials $$G_{m}(x)$$ are given by the generating function

$$\frac{2t}{e^{t}+1}e^{xt} =\sum _{m=0}^{\infty}G_{m}(x) \frac{t^{m}}{m!} \quad(\mbox{see [1--12]}).$$
(1.1)

The first few Genocchi polynomials are as follows:

\begin{aligned}&G_{0}(x)=0,\qquad G_{1}(x)=1,\qquad G_{2}(x)=2x-1, \\ &G_{3}(x)=3x^{2}-3x,\qquad G_{4}(x)=4x^{3}-6x^{2}+1, \\ &G_{5}(x)=5x^{4}-10x^{3}+5x, \qquad G_{6}(x)=6x^{5}-15x^{4}+15x^{2}-3. \end{aligned}
(1.2)

From the relation $$G_{m}(x)=mE_{m-1}(x) (m\geq1)$$, the following facts are obtained:

\begin{aligned}& \deg G_{m}(x)=m-1\quad (m\geq1), \qquad G_{m}=mE_{m-1}\quad (m\geq1), \\ & G_{0}=0, \qquad G_{1} =1,\qquad G_{2m+1}=0\quad (m \geq1),\quad \mbox{and}\quad G_{2m}\neq 0\quad (m\geq1). \end{aligned}
(1.3)

\begin{aligned} & \frac{d}{dx} G_{m}(x)=mG_{m-1}(x) \quad (m\geq1), \\ & G_{m}(x+1)+G_{m}(x)=2mx^{m-1}\quad (m\geq0). \end{aligned}
(1.4)

From these, we immediately obtain

\begin{aligned} G_{m}(1)+G_{m}(0)=2 \delta_{m,1} \quad (m\geq0) \end{aligned}
(1.5)

and

\begin{aligned} \int_{0}^{1} G_{m}(x) \,dx &= \frac{1}{m+1} \bigl(G_{m+1}(1)-G_{m+1}(0) \bigr) \\ &= \frac{2}{m+1} \bigl(-G_{m+1}(0)+\delta_{m,0} \bigr) \\ & = \textstyle\begin{cases} 0& \text{if } m \textit{ is even}, \\ -\frac{2}{m+1}G_{m+1} & \text{if } m \textit{ is odd}. \end{cases}\displaystyle \end{aligned}
(1.6)

For any real number x, we let $$\langle x\rangle =x-[x]\in[0,1)$$ denote the fractional part of x.

In this paper, we will consider three types of sums of finite products of Genocchi functions and derive the Fourier series expansions for them. Moreover, we will be able to express each of them in terms of Bernoulli functions $$B_{m}(\langle x\rangle )$$:

1. (1)

$$\alpha_{m}(\langle x\rangle )=\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} G_{i_{1}}(\langle x\rangle )\cdots G_{i_{r}}(\langle x\rangle ) (m>r\geq1)$$;

2. (2)

$$\beta_{m}(\langle x\rangle )= \sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} \frac{1}{i_{1}!\cdots i_{r}!}G_{i_{1}}(\langle x\rangle ) \cdots G_{i_{r}}(\langle x\rangle ) (m>r\geq1)$$;

3. (3)

$$\gamma_{m}(\langle x\rangle )=\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots, i_{r}\geq1} \frac{1}{i_{1}\cdots i_{r}}G_{i_{1}}(\langle x\rangle )\cdots G_{i_{r}}(\langle x\rangle ) (m>r\geq1)$$.

For elementary facts about Fourier analysis, the reader may refer to any book (for example, see [13, 14]).

As to $$\gamma_{m}(\langle x\rangle )$$, we note that the polynomial identity (1.7) follows immediately from Theorems 4.1 and 4.2, which are in turn derived from the Fourier series expansion of $$\gamma_{m}(\langle x\rangle )$$.

\begin{aligned} & \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1} \frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}}(x)G_{i_{2}}(x)\cdots G_{i_{r}}(x) \\ &\quad =\frac{1}{m} \Lambda_{m+1} + \frac{1}{m} \sum _{j=1}^{m-r}\binom{m}{j} \Lambda_{m-j+1}B_{j}(x), \end{aligned}
(1.7)

where, for $$l>r$$,

\begin{aligned} \Lambda_{l} ={}& \sum _{0\leq a \leq r}\binom{r}{a} (-1)^{a} 2^{r-a} \sum_{i_{1}+\cdots+i_{a}=l+a-r, i_{1},\ldots,i_{a} \geq1} \frac{G_{i_{1}}G_{i_{2}}\cdots G_{i_{a}}}{i_{1}i_{2} \cdots i_{a}} \\ &{} -\sum_{i_{1}+\cdots+i_{r}=l, i_{1},\ldots,i_{r}\geq1} \frac{G_{i_{1}}G_{i_{2}}\cdots G_{i_{r}}}{i_{1}i_{2} \cdots i_{r}}. \end{aligned}
(1.8)

The obvious polynomial identities can be derived also for $$\alpha _{m}(\langle x\rangle )$$ and $$\beta_{m}(\langle x\rangle )$$ from Theorems 2.1 and 2.2, and Theorems 3.1 and 3.2, respectively. It is noteworthy that from the Fourier series expansion of the function

$$\sum_{k=1}^{m-1} \frac{1}{k(m-k)}B_{k} \bigl( \langle x \rangle \bigr)B_{m-k} \bigl( \langle x \rangle \bigr)$$
(1.9)

we can derive the Faber-Pandharipande-Zagier identity (see [1, 1520]) and the Miki identity (see [1723]). In case of $$r=2$$, $$\gamma_{m}(\langle x\rangle )=\sum_{k=1}^{m-1} \frac{1}{k(m-k)}G_{k}( \langle x \rangle )G_{m-k}( \langle x \rangle )$$, and hence our problem here is a natural extension of the previous work, which leads to a simple proof for the important Faber-Pandharipande-Zagier and Miki identities (see [16, 22]). We will give an outline below, and this may be viewed as the main motivation for the present study.

The following polynomial identity follows immediately from the Fourier series expansion of the function in (1.9):

\begin{aligned} & \sum_{k=1}^{m-1} \frac{1}{k (m-k )}B_{k} (x )B_{m-k} (x ) \\ &\quad = \frac{2}{m^{2}} \biggl(B_{m}+\frac{1}{2} \biggr)+ \frac{2}{m}\sum_{k=1}^{m-2} \frac{1}{m-k}\dbinom{m}{k}B_{m-k}B_{k} (x )+ \frac{2}{m}H_{m-1}B_{m} (x ) \quad (m\geq2), \end{aligned}
(1.10)

where $$H_{m}=\sum_{j=1}^{m}\frac{1}{j}$$ are the harmonic numbers.

Simple modification of (1.10) yields

\begin{aligned} & \sum_{k=1}^{m-1} \frac{1}{2k (2m-2k )}B_{2k} (x )B_{2m-2k} (x )+\frac{2}{2m-1}B_{1} (x )B_{2m-1} (x ) \\ &\quad = \frac{1}{m}\sum_{k=1}^{m} \frac{1}{2k}\dbinom {2m}{2k}B_{2k}B_{2m-2k} (x )+ \frac{1}{m}H_{2m-1}B_{2m} (x ) \\ &\qquad{}+\frac{2}{2m-1}B_{1} (x )B_{2m-1}\quad (m\ge2 ). \end{aligned}
(1.11)

Letting $$x=0$$ in (1.11) gives a slightly different version of the well-known Miki identity (see [22]):

\begin{aligned} & \sum_{k=1}^{m-1} \frac{1}{2k (2m-2k )}B_{2k}B_{2m-2k} \\ &\quad = \frac{1}{m}\sum_{k=1}^{m} \frac{1}{2k}\dbinom {2m}{2k}B_{2k}B_{2m-2k}+ \frac{1}{m}H_{2m-1}B_{2m} \quad (m\ge 2 ). \end{aligned}
(1.12)

Setting $$x=\frac{1}{2}$$ in (1.12) with $$\overline{B}_{m}= (\frac{1-2^{m-1}}{2^{m-1}} )B_{m}= (2^{1-m}-1 )B_{m}=B_{m} (\frac{1}{2} )$$, we have

\begin{aligned} & \sum_{k=1}^{m-1} \frac{1}{2k (2m-2k )} \overline {B}_{2k}\overline{B}_{2m-2k} \\ &\quad = \frac{1}{m}\sum_{k=1}^{m} \frac{1}{2k}\dbinom {2m}{2k}B_{2k}\overline{B}_{2m-2k}+ \frac{1}{m}H_{2m-1}\overline {B}_{2m}\quad (m\ge2 ), \end{aligned}
(1.13)

which is the Faber-Pandharipande-Zagier identity (see [16]). Some of the different proofs of Miki’s identity can be found in [15, 2123]. Miki in [22] exploits a formula for the Fermat quotient $$\frac{a^{p}-a}{p}$$ modulo $$p^{2}$$, Shiratani-Yokoyama in [23] employs p-adic analysis, Gessel in [21] bases his work on two different expressions for Stirling numbers of the second kind $$S_{2} (n,k )$$, and Dunne-Schubert in [15] uses the asymptotic expansion of some special polynomials coming from the quantum field theory computations. As we can see, all of these proofs are quite involved. On the other hand, our proof of Miki’s and Faber-Pandharipande-Zagier’s identities follow from the polynomial identity (1.10), which in turn follows immediately from the Fourier series expansion of (1.9), together with the elementary manipulations outlined in (1.11)-(1.13). Some related recent work can be found in [10, 2426].

## 2 The first type of sums of finite products

In this section, we will derive the Fourier series of the first type of sums of products of Genocchi functions. Let us denote

$$\alpha_{m}(x)=\sum _{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} G_{i_{1}}(x)G_{i_{2}}(x)\cdots G_{i_{r}}(x)\quad (m>r\geq1).$$
(2.1)

Here the sum runs over all positive integers $$i_{1}, \ldots, i_{r}$$ with $$i_{1}+\cdots+i_{r}=m, (m>r\geq1)$$. Note here that $$\deg\alpha_{m} (x)=m-r\geq1$$. Then we will consider the function

$$\alpha_{m} \bigl(\langle x \rangle \bigr)= \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} G_{i_{1}} \bigl( \langle x\rangle \bigr)G_{i_{2}} \bigl(\langle x\rangle \bigr)\cdots G_{i_{r}} \bigl( \langle x\rangle \bigr)\quad (m>r\geq1),$$
(2.2)

defined on $$\mathbb{R}$$, which is periodic with period 1. The Fourier series of $$\alpha_{m}(\langle x\rangle )$$ is

$$\sum_{n=-\infty}^{\infty}A_{n}^{(m)} e^{2\pi i n x},$$
(2.3)

where

$$A_{n}^{(m)}= \int_{0}^{1} \alpha_{m} \bigl(\langle x \rangle \bigr) e^{-2\pi inx} \,dx= \int_{0}^{1} \alpha _{m} (x) e^{-2\pi inx} \,dx.$$
(2.4)

Before proceeding, we note the following:

\begin{aligned} \alpha_{m}'(x)={}& \sum _{i_{1}+\cdots+i_{r}=m, i_{1}\geq 2, i_{2},\ldots ,i_{r}\geq1} i_{1}G_{i_{1}-1}(x)G_{i_{2}}(x) \cdots G_{i_{r}}(x) \\ &{} +\cdots+ \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r-1}\geq 1,i_{r}\geq2} i_{r} G_{i_{1}}(x)\cdots G_{i_{r-1}}G_{{i_{r}}-1}(x) \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots,i_{r}\geq1} (i_{1}+1)G_{i_{1}}(x)G_{i_{2}}(x) \cdots G_{i_{r}}(x) \\ &{} +\cdots+ \sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots,i_{r}\geq 1}(i_{r}+1) G_{i_{1}}(x)G_{i_{2}}(x)\cdots G_{i_{r}}(x) \\ ={}&(m+r-1)\sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots,i_{r}\geq1} G_{i_{1}}(x)G_{i_{2}}(x) \cdots G_{i_{r}}(x) \\ ={}&(m+r-1) \alpha_{m-1}(x). \end{aligned}
(2.5)

So, $$\alpha_{m}'(x)=(m+r-1)\alpha_{m-1}(x)$$, and from this, we obtain

$$\biggl( \frac{\alpha_{m+1}(x)}{m+r} \biggr)' = \alpha_{m}(x)$$
(2.6)

and

$$\int_{0}^{1} \alpha_{m}(x) \,dx = \frac{1}{m+r} \bigl(\alpha_{m+1}(1)-\alpha_{m+1}(0) \bigr).$$
(2.7)

We put $$\Delta_{m}=\alpha_{m}(1)- \alpha_{m}(0)$$, for $$m>r$$. Then we have

\begin{aligned} \Delta_{m} ={}& \alpha_{m}(1)- \alpha_{m}(0) \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} \bigl(G_{i_{1}}(1)G_{i_{2}}(1) \cdots G_{i_{r}}(1) - G_{i_{1}}G_{i_{2}}\cdots G_{i_{r}} \bigr) \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} \bigl( (-G_{i_{1}} +2 \delta_{i_{1},1} )\cdots (-G_{i_{r}}+2 \delta _{i_{r},1} ) -G_{i_{1}}\cdots G_{i_{r}} \bigr) \\ ={}&\sum_{0\leq a \leq r} \binom{r}{a} \sum _{i_{1}+\cdots+i_{a}=m+a-r, i_{1}, \ldots,i_{a}\geq1}(-1)^{a} 2^{r-a} G_{i_{1}} \cdots G_{i_{a}} \\ &{} - \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} G_{i_{1}}G_{i_{2}} \cdots G_{i_{r}}, \end{aligned}
(2.8)

where we understand that, for $$a=0$$, the inner sum is $$2^{r} \delta _{m,r}$$. Observe here that the sums over all $$i_{1}+\cdots+i_{r}=m\ (i_{1}, \ldots, i_{r}\geq1)$$ of any term with a of $$-G_{i_{e}}$$ and b of $$2\delta_{i_{f},1}\ (1\leq e, f\leq r, a+b=r)$$ all give

\begin{aligned} &\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} (-G_{i_{1}} ) \cdots (-G_{i_{a}} ) (2\delta _{i_{a+1},1}) \cdots(2\delta_{i_{a+b},1}) \\ &\quad=\sum_{i_{1}+\cdots+i_{a}=m+a-r, i_{1},\ldots, i_{a}\geq1} (-1)^{a} 2^{r-a} G_{i_{1}}\cdots G_{i_{a}}. \end{aligned}
(2.9)

Note that, as $$i_{1}+\cdots+i_{a}=m+a-r>a$$, the above sum is not empty. From the definition of $$\Delta_{m}$$, we have

\begin{aligned} \begin{aligned} &\alpha_{m}(0)= \alpha_{m}(1) \quad\Longleftrightarrow\quad \Delta_{m} =0, \\ & \int_{0}^{1} \alpha_{m}(x) \,dx = \frac{1}{m+r}\Delta_{m+1}. \end{aligned} \end{aligned}
(2.10)

Now, we want to determine the Fourier coefficients $$A_{n}^{(m)}$$.

Case 1: $$n\neq0$$. We have

\begin{aligned} A_{n}^{(m)} &= \int_{0}^{1} \alpha_{m} (x) e^{-2\pi inx} \,dx \\ &= -\frac{1}{2\pi in } \bigl[ \alpha_{m} (x) e^{-2\pi inx} \bigr]_{0}^{1} + \frac{1}{2\pi in } \int_{0}^{1} \alpha_{m}' (x) e^{-2\pi inx} \,dx \\ &= -\frac{1}{2\pi in } \bigl(\alpha_{m} (1)-\alpha_{m} (0) \bigr)+ \frac {m+r-1}{2\pi in} \int_{0}^{1} \alpha_{m-1}(x) e^{-2\pi in x}\,dx \\ &= \frac{m+r-1}{2\pi in} A_{n}^{(m-1)}- \frac{1}{2\pi in} \Delta_{m} \\ &=\frac{m+r-1}{2\pi in } \biggl(\frac{m+r-2}{2\pi in} A_{n}^{(m-2)}- \frac{1}{2\pi in} \Delta_{m-1} \biggr) - \frac{1}{2\pi in} \Delta_{m} \\ &=\frac{(m+r-1)_{2}}{(2\pi in)^{2} } A_{n}^{(m-2)} - \sum _{j=1}^{2} \frac {(m+r-1)_{j-1}}{(2\pi in)^{j} } \Delta_{m-j+1} \\ &= \cdots \\ &=\frac{(m+r-1)_{m-r}}{(2\pi in)^{m-r} } A_{n}^{(r)} - \sum _{j=1}^{m-r} \frac{(m+r-1)_{j-1}}{(2\pi in)^{j} } \Delta_{m-j+1} \\ &=-\frac{1}{m+r}\sum_{j=1}^{m-r} \frac{(m+r)_{j}}{(2\pi in)^{j}}\Delta_{m-j+1}, \end{aligned}
(2.11)

where

$$A_{n}^{(r)}= \int_{0}^{1} e^{-2\pi in x} \,dx=0.$$
(2.12)

Case 2: $$n=0$$. We have

$$A_{0}^{(m)} = \int_{0}^{1} \alpha_{m}(x) \,dx= \frac{1}{m+r}\Delta_{m+1}.$$
(2.13)

We recall the following facts about Bernoulli functions $$B_{m}(\langle x\rangle )$$:

1. (a)

for $$m\geq2$$,

\begin{aligned} B_{m } \bigl(\langle x\rangle \bigr) = -m! \sum_{n=-\infty, n\neq0}^{\infty}\frac{e^{2\pi inx}}{(2\pi in)^{m}}, \end{aligned}
(2.14)
2. (b)

for $$m=1$$,

\begin{aligned} -\sum_{n=-\infty, n\neq0}^{\infty}\frac{e^{2\pi inx}}{2\pi in} = \textstyle\begin{cases} B_{1}(\langle x\rangle ) & \text{for } x \in\mathbb{Z}^{c}, \\ 0 & \text{for } x \in\mathbb{Z}, \end{cases}\displaystyle \end{aligned}
(2.15)

where $$\mathbb{Z}^{c}=\mathbb{R}-\mathbb{Z}$$. $$\alpha_{m}(\langle x\rangle )\ (m>r\geq1)$$ is piecewise $$C^{\infty}$$. Moreover, $$\alpha_{m}(\langle x\rangle )$$ is continuous for those positive integers $$m>r$$ with $$\Delta_{m}=0$$ and discontinuous with jump discontinuities at integers for those positive integers $$m>r$$ with $$\Delta_{m}\neq0$$. Assume first that $$\Delta_{m}=0$$, for a positive integer $$m>r$$. Then $$\alpha_{m}(0)=\alpha_{m}(1)$$. Hence $$\alpha_{m}(\langle x\rangle )$$ is piecewise $$C^{\infty}$$, and continuous. Thus, the Fourier series of $$\alpha_{m}(\langle x\rangle )$$ converges uniformly to $$\alpha_{m}(\langle x\rangle )$$, and

\begin{aligned} &\alpha_{m} \bigl(\langle x \rangle \bigr) \\ &\quad= \frac{1}{m+r}\Delta_{m+1} + \sum _{n=-\infty, n\neq0}^{\infty}\Biggl( -\frac{1}{m+r}\sum _{j=1}^{m-r} \frac{(m+r)_{j}}{(2\pi in)^{j}} \Delta_{m-j+1} \Biggr)e^{2\pi in x} \\ &\quad= \frac{1}{m+r}\Delta_{m+1} + \frac{1}{m+r} \sum _{j=1}^{m-r}\binom {m+r}{j} \Delta_{m-j+1} \Biggl( -j! \sum_{n=-\infty, n\neq0}^{\infty}\frac{e^{2\pi i nx}}{(2\pi in)^{j}} \Biggr) \\ &\quad= \frac{1}{m+r}\Delta_{m+1} + \frac{1}{m+r} \sum _{j=2}^{m-r}\binom {m+r}{j} \Delta_{m-j+1} B_{j} \bigl(\langle x\rangle \bigr) \\ &\qquad{} + \Delta_{m} \times \textstyle\begin{cases} B_{1}(\langle x\rangle ) & \text{for } x \in\mathbb{Z}^{c}, \\ 0 & \text{for }x \in\mathbb{Z}. \end{cases}\displaystyle \end{aligned}
(2.16)

Now, we can state our first result.

### Theorem 2.1

For each positive integer l, with $$l>r$$, we let

\begin{aligned} \Delta_{l} ={}& \sum _{0\leq a \leq r} \binom{r}{a} \sum_{i_{1}+\cdots +i_{a}=l+a-r, i_{1},\ldots,i_{a}\geq1} (-1)^{a}2^{r-a}G_{i_{1}}\cdots G_{i_{a}} \\ &{} - \sum_{i_{1}+\cdots+i_{r}=l, i_{1},\ldots,i_{r} \geq1} G_{i_{1}}\cdots G_{i_{r}}. \end{aligned}
(2.17)

Assume that $$\Delta_{m}=0$$, for a positive integer $$m> r$$. Then we have the following:

1. (a)

$$\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1}G_{i_{1}}(\langle x\rangle ) \cdots G_{i_{r}}(\langle x\rangle )$$ has the Fourier series expansion

\begin{aligned} &\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1}G_{i_{1}} \bigl(\langle x\rangle \bigr) \cdots G_{i_{r}} \bigl(\langle x\rangle \bigr) \\ &\quad= \frac{1}{m+r}\Delta_{m+1} + \sum _{n=-\infty, n\neq0}^{\infty}\Biggl(- \frac{1}{m+r} \sum _{j=1}^{m-r} \frac{(m+r)_{j}}{(2\pi in)^{j}} \Delta_{m-j+1} \Biggr)e^{2\pi in x}, \end{aligned}
(2.18)

for all $$x\in\mathbb{R}$$, where the convergence is uniform,

2. (b)
\begin{aligned} &\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1}G_{i_{1}} \bigl(\langle x\rangle \bigr) \cdots G_{i_{r}} \bigl(\langle x\rangle \bigr) \\ &\quad = \frac{1}{m+r}\Delta_{m+1} +\frac{1}{m+r} \sum _{j=2}^{m-r} \binom {m+r}{j} \Delta_{m-j+1}B_{j} \bigl(\langle x\rangle \bigr), \end{aligned}
(2.19)

for all $$x\in\mathbb{R}$$, where $$B_{j}(\langle x\rangle )$$ is the Bernoulli function.

Assume next that $$\Delta_{m}\neq0$$, for a positive integer $$m>r$$. Then $$\alpha_{m}(0)\neq\alpha_{m}(1)$$. Hence $$\alpha_{m}(\langle x\rangle )$$ is piecewise $$C^{\infty}$$, and discontinuous with jump discontinuities at integers. The Fourier series of $$\alpha _{m}(\langle x\rangle )$$ converges pointwise to $$\alpha_{m}(\langle x\rangle )$$, for $$x \in\mathbb{Z}^{c}$$, and it converges to

$$\frac{1}{2} \bigl( \alpha_{m}(0)+ \alpha_{m}(1) \bigr)= \alpha_{m}(0)+ \frac {1}{2} \Delta_{m},$$
(2.20)

for $$x\in\mathbb{Z}$$. Now, we can state our second result.

### Theorem 2.2

For each positive integer l, with $$l>r$$, we let

\begin{aligned} \Delta_{l} ={}& \sum _{0\leq a\leq r } \binom{r}{a} \sum_{i_{1}+\cdots +i_{a}=l+a-r, i_{1},\ldots,i_{a}\geq1} (-1)^{a}2^{r-a} G_{i_{1}} \cdots G_{i_{a}} \\ &{} - \sum_{i_{1}+\cdots+i_{r}=l, i_{1},\ldots,i_{r}\geq1} G_{i_{1}} \cdots G_{i_{r}}. \end{aligned}
(2.21)

Assume that $$\Delta_{m} \neq0$$, for a positive integer $$m> r$$. Then we have the following:

1. (a)
\begin{aligned} & \frac{1}{m+r}\Delta_{m+1} + \sum _{n=-\infty, n\neq0}^{\infty}\Biggl(-\frac{1}{m+r} \sum _{j=1}^{m-r} \frac{(m+r)_{j}}{(2\pi in)^{j}} \Delta_{m-j+1} \Biggr) e^{2\pi inx} \\ & \quad= \textstyle\begin{cases} \sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} G_{i_{1}}(\langle x\rangle ) \cdots G_{i_{r}}(\langle x\rangle ) & \textit{for } x \in\mathbb{Z}^{c}, \\ \sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} G_{i_{1}} \cdots G_{i_{r}} +\frac{1}{2}\Delta_{m} & \textit{for }x \in\mathbb{Z}, \end{cases}\displaystyle \end{aligned}
(2.22)
2. (b)
\begin{aligned} \begin{aligned} & \frac{1}{m+r} \Delta_{m+1} + \frac{1}{m+r} \sum_{j=1}^{m-r} \binom{m+r}{j} \Delta_{m-j+1} B_{j} \bigl(\langle x\rangle \bigr) \\ &\quad = \sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} G_{i_{1}} \bigl(\langle x \rangle \bigr) \cdots G_{i_{r}} \bigl(\langle x\rangle \bigr), \quad \textit{for } x\in \mathbb{Z}^{c}, \end{aligned} \end{aligned}
(2.23)
\begin{aligned} \begin{aligned} & \frac{1}{m+r} \Delta_{m+1} + \frac{1}{m+r} \sum_{j=2}^{m-r} \binom{m+r}{j} \Delta_{m-j+1} B_{j} \bigl(\langle x\rangle \bigr) \\ &\quad=\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} G_{i_{1}} \cdots G_{i_{r}}+ \frac{1}{2}\Delta_{m}, \quad x\in\mathbb{Z}. \end{aligned} \end{aligned}
(2.24)

## 3 The second type of sums of finite products

Let $$\beta_{m}(x)=\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1}\frac {1}{i_{1}!\cdots i_{r}!} G_{i_{1}}(x) \cdots G_{i_{r}}(x)\ (m>r\geq1)$$. Here the sum runs over all positive integers $$i_{1}, \ldots, i_{r}$$ with $$i_{1}+\cdots+i_{r}=m\ (r\geq1)$$. Then we will consider the function

$$\beta_{m} \bigl(\langle x\rangle \bigr)= \sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} \frac {1}{i_{1}!\cdots i_{r}!}G_{i_{1}} \bigl( \langle x\rangle \bigr) \cdots G_{i_{r}} \bigl(\langle x\rangle \bigr),$$
(3.1)

defined on $$\mathbb{R}$$, which is periodic with period 1. The Fourier series of $$\beta_{m}(\langle x\rangle )$$ is

$$\sum_{n=-\infty}^{\infty}B_{n}^{(m)} e^{2\pi inx},$$
(3.2)

where

$$B_{n}^{(m)} = \int_{0}^{1} \beta_{m} \bigl(\langle x \rangle \bigr) e^{-2\pi inx} \,dx= \int_{0}^{1} \beta _{m}(x) e^{-2\pi inx} \,dx.$$
(3.3)

Before proceeding, we need to observe the following:

\begin{aligned} \beta_{m}'(x)={}& \sum _{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} \biggl\{ \frac{i_{1}}{i_{1}!\cdots i_{r}!}G_{i_{1}-1}(x)G_{i_{2}}(x) \cdots G_{i_{r}}(x) +\cdots \\ &{}+\frac{i_{r}}{i_{1}!\cdots i_{r}!}G_{i_{1}}(x) \cdots G_{i_{r-1}}(x)G_{i_{r}-1}(x) \biggr\} \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m, i_{1}\geq2, i_{2} ,\ldots, i_{r}\geq1} \frac {1}{(i_{1}-1)!i_{2}!\cdots i_{r}!}G_{i_{1}-1}(x) G_{i_{2}}(x)\cdots G_{i_{r}}(x) \\ &{} +\cdots+\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, i_{2} ,\ldots, i_{r-1} \geq1, i_{r}\geq2}\frac{1}{i_{1}!\cdots i_{r-1}!(i_{r}-1)!} G_{i_{1}}(x) \cdots G_{i_{r-1}}(x)G_{i_{r}-1}(x) \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots,i_{r}\geq1} \frac {1}{i_{1}!\cdots i_{r}!}G_{i_{1}}(x) \cdots G_{i_{r}}(x) \\ &{} +\cdots+\sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots, i_{r}\geq1} \frac{1}{i_{1}!\cdots i_{r}!}G_{i_{1}}(x) \cdots G_{i_{r}}(x) \\ ={}& r \beta_{m-1} (x). \end{aligned}
(3.4)

Thus $$\beta_{m}'(x) = r \beta_{m-1} (x)$$, and, from this, we obtain

$$\biggl( \frac{\beta_{m+1}(x)}{r} \biggr)' = \beta_{m}(x)$$
(3.5)

and

$$\int_{0}^{1} \beta_{m} (x) \,dx= \frac{1}{r} \bigl(\beta_{m+1}(1)- \beta_{m+1}(0) \bigr).$$
(3.6)

For $$m>r$$, let

\begin{aligned} \Omega_{m}={}& \beta_{m}(1)- \beta_{m}(0) \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} \frac {1}{i_{1}!\cdots i_{r}!}G_{i_{1}}(1) \cdots G_{i_{r}}(1) \\ &{} -\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1} \frac {1}{i_{1}!\cdots i_{r}!}G_{i_{1}} \cdots G_{i_{r}} \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} \frac {1}{i_{1}!\cdots i_{r}!}(-G_{i_{1}}+2 \delta_{i_{1},1}) \cdots(- G_{i_{r}}+2\delta_{i_{r},1}) \\ & {}-\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1} \frac {1}{i_{1}!\cdots i_{r}!}G_{i_{1}} \cdots G_{i_{r}} \\ ={}& \sum_{0\leq a\leq r} \binom{r}{a} \sum _{i_{1}+\cdots+i_{a}=m+a-r, i_{1}, \ldots,i_{a}\geq1}(-1)^{a} 2^{r-a} \frac{G_{i_{1}} G_{i_{2}}\cdots G_{i_{a}} }{ i_{1}!\cdots i_{a}!} \\ &{} -\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1} \frac {1}{i_{1}!\cdots i_{r}!}G_{i_{1}} G_{i_{2}}\cdots G_{i_{r}}, \end{aligned}
(3.7)

where we understand that, for $$a=0$$, the inner sum is $$2^{r}\delta_{m,r}$$. Observe that the sums over all $$i_{1}+ \cdots+i_{r}=m\ (i_{1}, \ldots i_{r}\geq 1)$$ of any term with a of $$-G_{i_{e}}$$ and b of $$2\delta_{i_{f},1}\ (1\leq e,f \leq r, a+b=r)$$ all give

\begin{aligned} &\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots,i_{r}\geq1} \frac{1}{i_{1}!\cdots i_{r}!}(-G_{i_{1}}) \cdots(-G_{i_{a}}) (2 \delta_{i_{a+1},1})\cdots(2\delta _{i_{a+b},1}) \\ &\quad= \sum_{i_{1}+\cdots+i_{a}=m+a-r, i_{1}, \ldots,i_{a}\geq1} \frac{ (-1)^{a} 2^{r-a}}{i_{1}!\cdots i_{a}!} G_{i_{1}} \cdots G_{i_{a}}. \end{aligned}
(3.8)

From the definition of $$\Omega_{m}$$, we have

\begin{aligned} &\beta_{m}(0)= \beta_{m}(1) \quad\Longleftrightarrow \quad \Omega_{m} =0, \\ & \int_{0}^{1} \beta_{m}(x) \,dx = \frac{1}{r}\Omega_{m+1}. \end{aligned}
(3.9)

Next, we want to determine the Fourier coefficients $$B_{n}^{(m)}$$.

Case 1: $$n \neq0$$. We have

\begin{aligned} B_{n}^{(m)} ={}& \int_{0}^{1} \beta_{m} (x) e^{-2\pi inx} \,dx \\ ={}&- \frac{1}{2 \pi in} \bigl[\beta_{m}(x)e^{-2 \pi inx} \bigr]_{0}^{1} + \frac{1}{2 \pi in} \int_{0}^{1} \beta_{m}' (x) e^{-2\pi inx} \,dx \\ ={}&{-} \frac{1}{2 \pi in} \bigl(\beta_{m}(1)- \beta_{m}(0) \bigr) \\ &{} + \frac{r}{ 2\pi in} \int_{0}^{1} \beta_{m-1} (x) e^{-2\pi inx} \,dx \\ ={}& \frac{r}{ 2\pi in}B_{n}^{(m-1)}- \frac{1}{2\pi in} \Omega_{m} \\ ={}& \frac{r}{ 2\pi in} \biggl( \frac{r}{ 2\pi in}B_{n}^{(m-2)} - \frac {1}{2\pi in}\Omega_{m-1} \biggr) - \frac{1}{2\pi in} \Omega_{m} \\ ={}& \biggl(\frac{r}{2 \pi in} \biggr)^{2}B_{n}^{(m-2)} -\sum_{j=1}^{2} \frac {r^{j-1}}{(2\pi in)^{j}} \Omega_{m-j+1} \\ ={}& \cdots \\ ={} &\biggl( \frac{r}{ 2\pi in} \biggr)^{m-r}B_{n}^{(r)}- \sum_{j=1}^{m-r} \frac{r^{j-1}}{(2\pi in)^{j}} \Omega_{m-j+1} \\ ={} &- \sum_{j=1}^{m-r} \frac{r^{j-1}}{(2\pi in)^{j}} \Omega_{m-j+1}, \end{aligned}
(3.10)

where

$$B_{n}^{(r)}= \int_{0}^{1} e^{-2\pi inx} \,dx=0.$$
(3.11)

Case 2: $$n=0$$. We have

$$B_{0}^{(m)}= \int_{0}^{1} \beta_{m}(x) \,dx= \frac{1}{r}\Omega_{m+1}.$$
(3.12)

$$\beta_{m}(\langle x\rangle )\ (m>r \geq1)$$ is piecewise $$C^{\infty}$$. Moreover, $$\beta_{m} (\langle x\rangle )$$ is continuous for those positive integers $$m>r$$ with $$\Omega_{m}=0$$ and discontinuous with jump discontinuities at integers for those integers $$m>r$$ with $$\Omega _{m}\neq0$$.

Assume first that $$\Omega_{m}=0$$, for a positive integer $$m>r$$. Then $$\beta_{m}(0)=\beta_{m}(1)$$. Hence $$\beta_{m}(\langle x\rangle )$$ is piecewise $$C^{\infty}$$, and continuous. Thus the Fourier series of $$\beta_{m}(\langle x\rangle )$$ converges uniformly to $$\beta_{m}(\langle x\rangle )$$, and

\begin{aligned} \beta_{m} \bigl(\langle x \rangle \bigr)&= \frac{1}{r} \Omega_{m+1} + \sum_{n=-\infty, n\neq 0}^{\infty}\Biggl(- \sum_{j=1}^{m-r} \frac{r^{j-1}}{(2\pi in)^{j}} \Omega_{m-j+1} \Biggr)e^{2 \pi in x} \\ &= \frac{1}{r} \Omega_{m+1} + \sum_{j=1}^{m-r} \frac{r^{j-1}}{j!} \Omega_{m-j+1} \Biggl( -j!\sum _{n=-\infty, n\neq0}^{\infty}\frac{e^{2 \pi inx} }{(2\pi in)^{j}} \Biggr) \\ &= \frac{1}{r} \Omega_{m+1} + \sum_{j=2}^{m-r} \frac{r^{j-1}}{j!} \Omega_{m-j+1}B_{j} \bigl(\langle x\rangle \bigr) +\Omega_{m} \times \textstyle\begin{cases} B_{1}(\langle x\rangle ) & \text{for } x \in\mathbb{Z}^{c}, \\ 0 & \text{for }x \in\mathbb{Z}. \end{cases}\displaystyle \end{aligned}
(3.13)

Now, we can state our first result.

### Theorem 3.1

For each positive integer l, with $$l>r$$, we let

\begin{aligned} \Omega_{l} ={}& \sum _{0\leq a \leq r} \binom{r}{a} (-1)^{a} 2^{r-a} \sum_{i_{1}+\cdots+i_{a}=l+a-r, i_{1},\ldots,i_{a}\geq1} \frac{G_{i_{1}}G_{i_{2}}\cdots G_{i_{a}}}{i_{1}!i_{2}!\cdots i_{a}!} \\ &{} - \sum_{i_{1}+\cdots+i_{r}=l, i_{1},\ldots, i_{r}\geq1} \frac{G_{i_{1}}G_{i_{2}}\cdots G_{i_{r}}}{i_{1}!i_{2}!\cdots i_{r}!}. \end{aligned}
(3.14)

Assume that $$\Omega_{m}=0$$, for a positive integer $$m>r$$. Then we have the following:

1. (a)

$$\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} \frac{1}{i_{1}! i_{2}!\cdots i_{r}} G_{i_{1}}(\langle x\rangle )G_{i_{2}}(\langle x\rangle )\cdots G_{i_{r}}(\langle x\rangle )$$ has the Fourier series expansion

\begin{aligned} &\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} \frac{1}{i_{1}! i_{2}!\cdots i_{r}!} G_{i_{1}} \bigl(\langle x\rangle \bigr)G_{i_{2}} \bigl(\langle x\rangle \bigr)\cdots G_{i_{r}} \bigl(\langle x\rangle \bigr) \\ &\quad = \frac{1}{r}\Omega_{m+1} + \sum _{n=-\infty, n\neq0}^{\infty}\Biggl(- \sum _{j=1}^{m-r} \frac{r^{j-1}}{(2\pi in)^{j}} \Omega_{m-j+1} \Biggr)e^{2\pi in x}, \end{aligned}
(3.15)

for all $$x\in\mathbb{R}$$, where the convergence is uniform,

2. (b)
\begin{aligned} &\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} \frac{1}{i_{1}! i_{2}!\cdots i_{r}} G_{i_{1}} \bigl(\langle x\rangle \bigr)G_{i_{2}} \bigl(\langle x\rangle \bigr)\cdots G_{i_{r}} \bigl(\langle x\rangle \bigr) \\ &\quad= \frac{1}{r}\Omega_{m+1} +\sum _{j=2}^{m-r} \frac{r^{j-1}}{j!} \Omega_{m-j+1}B_{j} \bigl(\langle x\rangle \bigr), \end{aligned}
(3.16)

for all $$x\in\mathbb{R}$$, where $$B_{j}(\langle x\rangle )$$ is the Bernoulli function.

Assume next that $$\Omega_{m} \neq0$$, for a positive integer $$m>r$$. Then $$\beta_{m}(0)\neq\beta_{m}(1)$$. Thus $$\beta_{m}(\langle x\rangle )$$ is piecewise $$C^{\infty}$$, and discontinuous with jump discontinuities at integers. The Fourier series of $$\beta_{m}(\langle x\rangle )$$ converges pointwise to $$\beta_{m}(\langle x\rangle )$$, for $$x \in\mathbb{Z}^{c}$$, and it converges to

\begin{aligned} \frac{1}{2} \bigl( \beta_{m}(0)+ \beta_{m}(1) \bigr)={}& \beta_{m}(0)+ \frac {1}{2} \Omega_{m} \\ ={}&\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots,i_{r}\geq1} \frac{1}{i_{1}! i_{2}!\cdots i_{r}} G_{i_{1}}G_{i_{2}} \cdots G_{i_{r}}+\frac{1}{2} \Omega_{m}, \end{aligned}
(3.17)

for $$x \in\mathbb{Z}$$. Now, we can state our second theorem.

### Theorem 3.2

For each positive integer l, with $$l>r$$, we let

\begin{aligned} \Omega_{l} ={}& \sum _{0\leq a \leq r} \binom{r}{a} (-1)^{a} 2^{r-a} \sum_{i_{1}+\cdots+i_{a}=l+a-r, i_{1},\ldots,i_{a}\geq1} \frac{G_{i_{1}}G_{i_{2}} \cdots G_{i_{a}}}{i_{1}!i_{2}!\cdots i_{a}!} \\ &{} - \sum_{i_{1}+\cdots+i_{r}=l, i_{1},\ldots, i_{r}\geq1} \frac{G_{i_{1}}G_{i_{2}} \cdots G_{i_{r}}}{i_{1}!i_{2}!\cdots i_{r}!}. \end{aligned}
(3.18)

Assume that $$\Omega_{m} \neq0$$, for a positive integer $$m>r$$. Then we have the following:

1. (a)
\begin{aligned} &\frac{1}{r}\Omega_{m+1} + \sum _{n=-\infty, n\neq0}^{\infty}\Biggl(-\sum _{j=1}^{m-r} \frac{r^{j-1}}{(2\pi in)^{j}} \Omega_{m-j+1} \Biggr) e^{2\pi inx} \\ &\quad = \textstyle\begin{cases} \sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots, i_{r}\geq1} \frac{1}{i_{1}!i_{2}!\cdots i_{r}!}G_{i_{1}}(\langle x\rangle )G_{i_{2}}(\langle x\rangle ) \cdots G_{i_{r}}(\langle x\rangle ) & \textit{for } x \in\mathbb{Z}^{c}, \\ \sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots, i_{r}\geq1} \frac{1}{i_{1}!i_{2}!\cdots i_{r}!}G_{i_{1}}G_{i_{2}}\cdots G_{i_{r}}+\frac {1}{2}\Omega_{m} & \textit{for } x \in\mathbb{Z}, \end{cases}\displaystyle \end{aligned}
(3.19)
2. (b)
\begin{aligned} & \frac{1}{r}\Omega_{m+1}+ \sum _{j=1}^{m-r} \frac{r^{j-1}}{j!} \Omega_{m-j+1}B_{j} \bigl(\langle x\rangle \bigr) \\ &\quad =\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots, i_{r}\geq1} \frac{1}{i_{1}!i_{2}!\cdots i_{r}!}G_{i_{1}} \bigl(\langle x\rangle \bigr)G_{i_{2}} \bigl(\langle x\rangle \bigr)\cdots G_{i_{r}} \bigl(\langle x\rangle \bigr), \end{aligned}
(3.20)

for $$x\in\mathbb{Z}^{c}$$, and

\begin{aligned} & \frac{1}{r}\Omega_{m+1}+ \sum _{j=2}^{m-r} \frac{r^{j-1}}{j!} \Omega_{m-j+1}B_{j} \bigl(\langle x\rangle \bigr) \\ & \quad =\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots, i_{r}\geq1} \frac{1}{i_{1}!i_{2}!\cdots i_{r}!}G_{i_{1}}G_{i_{2}} \cdots G_{i_{r}} + \frac {1}{2}\Omega_{m}, \end{aligned}
(3.21)

for $$x\in\mathbb{Z}$$.

## 4 The third type of sums of finite products

Let $$\gamma_{m}(x)=\sum_{i_{1}+\cdots+i_{r}=m, i_{1},\ldots, i_{r}\geq1} \frac{1}{i_{1}i_{2}\cdots i_{r}}G_{i_{1}}(x)\cdots G_{i_{r}}(x)\ (m>r\geq1)$$. Here the sum runs over all positive integers $$i_{1}, \ldots, i_{r}$$, with $$i_{1}+\cdots+i_{r}=m$$. Before proceeding, we observe the following:

\begin{aligned} \gamma_{m}'(x) ={}& \sum _{i_{1}+\cdots+i_{r}=m, i_{1}\geq2, i_{2},\ldots, i_{r}\geq1} \frac{1}{i_{2}\cdots i_{r}}G_{i_{1}-1}(x)G_{i_{2}}(x) \cdots G_{i_{r}} (x) \\ & {}+ \cdots \\ &{} + \sum_{i_{1}+\cdots+i_{r}=m, i_{r}\geq2, i_{1}, \ldots, i_{r-1}\geq1} \frac{1}{i_{1} \cdots i_{r-1}}G_{i_{1}}(x) \cdots G_{i_{r-1}}(x) G_{i_{r}-1} (x) \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots, i_{r}\geq1} \frac{1}{i_{2}\cdots i_{r}}G_{i_{1}}(x)G_{i_{2}}(x) \cdots G_{i_{r}} (x) \\ &{} +\cdots \\ & {}+ \sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots, i_{r}\geq1} \frac{1}{i_{1} \cdots i_{r-1}}G_{i_{1}}(x) \cdots G_{i_{r}} (x) \\ ={}&\sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots, i_{r}\geq1} \biggl\{ \frac{1}{i_{2}\cdots i_{r}}+ \frac{1}{i_{1}i_{3}\cdots i_{r}}+\cdots +\frac{1}{i_{1}\cdots i_{r-1}} \biggr\} G_{i_{1}}(x)G_{i_{2}}(x) \cdots G_{i_{r}} (x) \\ ={}& (m-1) \sum_{i_{1}+\cdots+i_{r}=m-1, i_{1}, \ldots, i_{r}\geq1}\frac {1}{i_{1}\cdots i_{r}} G_{i_{1}}(x)G_{i_{2}}(x)\cdots G_{i_{r}} (x) \\ ={}&(m-1)\gamma_{m-1}(x). \end{aligned}
(4.1)

So, $$\gamma_{m}'(x)=(m-1)\gamma_{m-1}(x)$$, and from this, we get

$$\biggl(\frac{\gamma_{m+1}(x)}{m} \biggr)'= \gamma_{m}(x)$$
(4.2)

and

\begin{aligned} & \int_{0}^{1} \gamma_{m}(x) \,dx= \frac{1}{m} \bigl(\gamma_{m+1}(1)-\gamma_{m+1}(0) \bigr). \end{aligned}
(4.3)

For $$m>r$$, we let

\begin{aligned} \Lambda_{m} ={}& \gamma_{m}(1)- \gamma_{m}(0) \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac {1}{i_{1}\cdots i_{r}} \bigl(G_{i_{1}}(1)\cdots G_{i_{r}} (1)-G_{i_{1}}\cdots G_{i_{r}} \bigr) \\ ={}& \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac {1}{i_{1}\cdots i_{r}} (-G_{i_{1}}+2 \delta_{i_{1},1})\cdots(-G_{i_{r}}+2 \delta_{i_{r},1}) \\ &{} - \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac {1}{i_{1}\cdots i_{r}} G_{i_{1}} \cdots G_{i_{r}} \\ ={}& \sum_{0\leq a\leq r} \binom{r}{a} (-1)^{a} 2^{r-a} \sum_{i_{1}+\cdots +i_{a}=m+a-r, i_{1}, \ldots, i_{a}\geq1} \frac{G_{i_{1}}G_{2}\cdots G_{i_{a}}}{i_{1}\cdots i_{a}} \\ &{} - \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac {1}{i_{1}\cdots i_{r}} G_{i_{1}} \cdots G_{i_{r}}, \end{aligned}
(4.4)

where we understand that, for $$a=0$$, the inner sum is $$2^{r}\delta _{m,r}$$. Note that

$$\gamma_{m}(0)= \gamma_{m}(1)\quad \Longleftrightarrow \quad \Lambda_{m}=0$$
(4.5)

and

\begin{aligned} \int_{0}^{1} \gamma_{m}(x) \,dx = \frac{1}{m} \Lambda_{m+1}. \end{aligned}
(4.6)

We are now going to consider the function

$$\gamma_{m} \bigl(\langle x \rangle \bigr)= \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac {1}{i_{1}\cdots i_{r}} G_{i_{1}} \bigl( \langle x\rangle \bigr)\cdots G_{i_{r}} \bigl(\langle x\rangle \bigr),$$
(4.7)

defined on $$\mathbb{R}$$, which is periodic with period 1. The Fourier series of $$\gamma_{m}(\langle x\rangle )$$ is

\begin{aligned} \sum_{n=-\infty}^{\infty}C_{n}^{(m)} e^{2\pi i n x}, \end{aligned}
(4.8)

where

$$C_{n}^{(m)} = \int_{0}^{1} \gamma_{m} \bigl(\langle x \rangle \bigr) e^{-2\pi inx} \,dx= \int_{0}^{1} \gamma _{m} (x) e^{-2\pi inx} \,dx.$$
(4.9)

Now, we would like to determine the Fourier coefficients $$C_{n}^{(m)}$$.

Case 1: $$n\neq0$$. We have

\begin{aligned} C_{n}^{(m)}&= -\frac{1}{2\pi in} \bigl[\gamma_{m}(x)e^{-2\pi in x} \bigr]_{0}^{1} + \frac{1}{2\pi in} \int_{0}^{1} \gamma_{m}'(x) e^{-2\pi in x} \,dx \\ &=- \frac{1}{2\pi in } \bigl(\gamma_{m}(1)-\gamma_{m}(0) \bigr) + \frac{m-1}{2\pi in } \int_{0}^{1} \gamma_{m-1}(x) e^{-2\pi in x}\,dx \\ &=\frac{m-1}{2\pi in }C_{n}^{(m-1)}- \frac{1}{2\pi in } \Lambda_{m} \\ &=\frac{m-1}{2\pi in } \biggl( \frac{m-2}{2\pi in}C_{n}^{(m-2)}- \frac {1}{2\pi in}\Lambda_{m-1} \biggr)- \frac{1}{2\pi in } \Lambda_{m} \\ &=\frac{(m-1)_{2}}{(2\pi in)^{2} }C_{n}^{(m-2)}-\sum _{j=1}^{2} \frac {(m-1)_{j-1}}{(2\pi in)^{j}}\Lambda_{m-j+1} \\ &= \cdots \\ &= \frac{(m-1)_{m-r}}{(2\pi in)^{m-r} }C_{n}^{(r)}-\sum _{j=1}^{m-r} \frac{(m-1)_{j-1}}{(2\pi in)^{j}}\Lambda_{m-j+1} \\ &=-\frac{1}{m}\sum_{j=1}^{m-r} \frac{(m)_{j}}{(2\pi in)^{j}}\Lambda_{m-j+1}. \end{aligned}
(4.10)

Note that

\begin{aligned} C_{n}^{(r)} = \int_{0}^{1} e^{-2 \pi inx}\,dx =0. \end{aligned}
(4.11)

Case 2: $$n=0$$. We have

\begin{aligned} C_{0}^{(m)} = \int_{0}^{1} \gamma_{m}(x) \,dx= \frac{1}{m} \Lambda_{m+1}. \end{aligned}
(4.12)

$$\gamma_{m}(\langle x\rangle ), (m>r \geq1)$$ is piecewise $$C^{\infty}$$. Moreover, $$\gamma_{m}(\langle x\rangle )$$ is continuous for those integers $$m>r$$ with $$\Lambda_{m} =0$$, and discontinuous with jump discontinuities at integers for those integers $$m>r$$ with $$\Lambda_{m} \neq0$$.

Assume first that $$\Lambda_{m}=0$$, for a positive integer $$m>r$$. Then $$\gamma_{m}(0)=\gamma_{m}(1)$$. Hence $$\gamma_{m}(\langle x\rangle )$$ is piecewise $$C^{\infty}$$, and continuous. So the Fourier series of $$\gamma_{m}(\langle x\rangle )$$ converges uniformly to $$\gamma_{m}(\langle x\rangle )$$, and

\begin{aligned} &\gamma_{m} \bigl(\langle x \rangle \bigr) \\ &\quad= \frac{1}{m} \Lambda_{m+1}+ \sum _{n=-\infty, n\neq0}^{\infty}\Biggl( - \frac{1}{m} \sum _{j=1}^{m-r} \frac{(m)_{j}}{(2\pi in)^{j}}\Lambda_{m-j+1} \Biggr)e^{2\pi inx} \\ &\quad=\frac{1}{m} \Lambda_{m+1}+ \frac{1}{m} \sum _{j=1}^{m-r}\binom {m}{j}\Lambda_{m-j+1} \Biggl( -j! \sum_{n=-\infty, n\neq0}^{\infty}\frac{e^{2\pi inx}}{(2\pi in)^{j}} \Biggr) \\ &\quad= \frac{1}{m} \Lambda_{m+1} + + \frac{1}{m} \sum _{j=2}^{m-r}\binom{m}{j} \Lambda_{m-j+1}B_{j} \bigl(\langle x\rangle \bigr) \\ &\qquad{} +\Lambda_{m} \times \textstyle\begin{cases} B_{1}(\langle x\rangle ) & \text{for } x \in\mathbb{Z}^{c}, \\ 0, & \text{for } x \in\mathbb{Z}. \end{cases}\displaystyle \end{aligned}
(4.13)

We are now ready to state our first theorem.

### Theorem 4.1

For each positive integer l, with $$l>r$$, we let

\begin{aligned} \Lambda_{l} ={}& \sum _{0\leq a \leq r } \binom{r}{a}(-1)^{a}2^{r-a} \sum_{i_{1}+\cdots+i_{a}=l+a-r, i_{1}, \ldots, i_{a}\geq1} \frac{G_{i_{1}}G_{i_{2}}\cdots G_{i_{a}}}{i_{1}i_{2}\cdots i_{a}} \\ &{} - \sum _{i_{1}+\cdots+i_{r}=l, i_{1}, \ldots, i_{r}\geq1} \frac{G_{i_{1}}G_{i_{2}}\cdots G_{i_{r}}}{i_{1}i_{2}\cdots i_{r}}. \end{aligned}
(4.14)

Assume that $$\Lambda_{m} =0$$, for a positive integer $$m>r$$. Then we have the following:

1. (a)

$$\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}}(\langle x\rangle )G_{i_{2}}(\langle x\rangle ) \cdots G_{i_{r}}(\langle x\rangle )$$ has the Fourier series expansion

\begin{aligned} &\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1} \frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}} \bigl(\langle x\rangle \bigr)G_{i_{2}} \bigl(\langle x\rangle \bigr)\cdots G_{i_{r}} \bigl(\langle x\rangle \bigr) \\ &\quad=\frac{1}{m} \Lambda_{m+1}+ \sum _{n=-\infty, n\neq0}^{\infty}\Biggl( - \frac{1}{m}\sum _{j=1}^{m-r} \frac{(m)_{j}}{(2\pi in)^{j}} \Lambda_{m-j+1} \Biggr) e^{2\pi in x}, \end{aligned}
(4.15)

for all $$x\in\mathbb{R}$$, where the convergence is uniform,

2. (b)

we have

\begin{aligned} &\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1} \frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}} \bigl(\langle x\rangle \bigr)G_{i_{2}} \bigl(\langle x\rangle \bigr)\cdots G_{i_{r}} \bigl(\langle x\rangle \bigr) \\ &\quad=\frac{1}{m} \Lambda_{m+1}+ \frac{1}{m}\sum _{j=2}^{m-r} \binom{m}{j} \Lambda_{m-j+1} B_{j} \bigl(\langle x\rangle \bigr), \end{aligned}
(4.16)

for all $$x\in\mathbb{R}$$, where $$B_{j}(\langle x\rangle )$$ is the Bernoulli function.

Assume next that $$\Lambda_{m}\neq0$$, for a positive integer $$m>r$$. Then $$\gamma_{m}(0)\neq\gamma_{m}(1)$$. Hence $$\gamma_{m}(\langle x\rangle )$$ is piecewise $$C^{\infty}$$, and discontinuous with jump discontinuities at integers. The Fourier series of $$\gamma _{m}(\langle x\rangle )$$ converges pointwise to $$\gamma_{m}(\langle x\rangle )$$, for $$x\in\mathbb{Z}^{c}$$, and it converges to

\begin{aligned} \frac{1}{2} \bigl( \gamma_{m}(0)+ \gamma_{m}(1) \bigr) &=\gamma_{m}(0)+ \frac{1}{2} \Lambda_{m} \\ &=\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}}G_{i_{2}} \cdots G_{i_{r}} + \frac{1}{2}\Lambda_{m}, \end{aligned}
(4.17)

for $$x\in\mathbb{Z}$$. Now, we state our second result.

### Theorem 4.2

For each positive integer l, with $$l>r$$, we let

\begin{aligned} \Lambda_{l} ={}& \sum _{0\leq a \leq r}\binom{r}{a} (-1)^{a} 2^{r-a} \sum_{i_{1}+\cdots+i_{a}=l+a-r, i_{1},\ldots,i_{a} \geq1} \frac{G_{i_{1}}G_{i_{2}} \cdots G_{i_{a}}}{i_{1}i_{2} \cdots i_{a}} \\ &{} -\sum_{i_{1}+\cdots+i_{r}=l, i_{1},\ldots,i_{r}\geq1} \frac{G_{i_{1}} G_{i_{2}}\cdots G_{i_{r}}}{i_{1}i_{2} \cdots i_{r}}. \end{aligned}
(4.18)

Assume that $$\Lambda_{m} \neq0$$, for a positive integer $$m>r$$. Then we have the following:

1. (a)
\begin{aligned} &\frac{1}{m} \Lambda_{m+1}+ \sum _{n=-\infty, n\neq0}^{\infty}\Biggl( - \frac{1}{m} \sum _{j=1}^{m-r}\frac{(m)_{j}}{(2\pi in)^{j}} \Lambda_{m-j+1} \Biggr) e^{2\pi in x} \\ &\quad = \textstyle\begin{cases} \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}}(\langle x\rangle ) G_{i_{2}}(\langle x\rangle ) \cdots G_{i_{r}}(\langle x\rangle )& \textit{for } x \in\mathbb{Z}^{c}, \\ \sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}}G_{i_{2}}\cdots G_{i_{r}} + \frac{1}{2}\Lambda_{m} & \textit{for } x \in\mathbb{Z}, \end{cases}\displaystyle \end{aligned}
(4.19)
2. (b)
\begin{aligned} & \frac{1}{m} \Lambda_{m+1} + \frac{1}{m} \sum_{j=1}^{m-r} \binom{m}{j} \Lambda_{m-j+1}B_{j} \bigl(\langle x\rangle \bigr) \\ &\quad =\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}} \bigl(\langle x\rangle \bigr)G_{i_{2}} \bigl(\langle x \rangle \bigr)\cdots G_{i_{r}} \bigl(\langle x\rangle \bigr), \end{aligned}
(4.20)

for $$x\in\mathbb{Z}^{c}$$, and

\begin{aligned} & \frac{1}{m} \Lambda_{m+1} + \frac{1}{m} \sum_{j=2}^{m-r} \binom {m}{j} \Lambda_{m-j+1} B_{j} \bigl(\langle x\rangle \bigr) \\ & \quad =\sum_{i_{1}+\cdots+i_{r}=m, i_{1}, \ldots, i_{r}\geq1}\frac{1}{i_{1} i_{2}\cdots i_{r}} G_{i_{1}}G_{i_{2}} \cdots G_{i_{r}} + \frac{1}{2} \Lambda_{m}, \end{aligned}
(4.21)

for $$x\in\mathbb{Z}$$.