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\begin{center}
\vskip 1cm{\LARGE\bf  Raabe's Identity and Covering Systems}
\vskip 1cm \large Yevgenya Movshovich \\ University of Illinois at Urbana-Champaign \\ Urbana, IL 61801 \\  USA \\
\href{mailto:yevgenya@illinois.edu}{yevgenya@illinois.edu}
\end{center}

\vskip .2 in

 \begin{abstract}  
A connection  between covering systems  and Bernoulli polynomials
established by Fraenkel, Beebee, and Porubsk\'y asserts that a function
$\sum_{i=1}^q \mu_i \chi_{A_i}(k)$ is an $M_A$-covering function for
a system of  arithmetic sequences $\{A_i\}_{i=1}^q$ if and only if it
satisfies a generalized Raabe identity. The connection is used to derive
recurrence formulas for the Bernoulli numbers.  Here we show that the
generalized Raabe identity  is a sum of Raabe's identities for $A_i$
multiplied by  $\mu_i$, and this sum is the direct source of the above
recurrence formulas. We show that many of these formulas are special
cases of the original multiplication formula of Raabe. We  find new
applications of the connection  to the covering systems.
\end{abstract}

\section{Introduction} \label{S1}
For $a\in \mathbb Z$ and $n\in \mathbb Z^+$ we let 
$
a(n) :=  a + n \mathbb Z =\{x\in \mathbb Z: x \equiv \modd{a} {n} \}.
$ 
Here $a(n)$ 
is called  an arithmetic sequence (with {\it common difference} $n$) or a residue class (with {\it modulus} $n$).
 Consider a system  $A=\{A_1, A_2, \ldots , A_q\}$ of  arithmetic sequences
  \begin{equation} \label{e1}
                   A_i=a_i(n_i)= a_i + n_i \mathbb Z,  \quad  0\le a_i&lt;n_i,  \quad i=1,\ldots, q, \quad q\ge 1.         
   \end{equation} 
  \begin{definition}  For  $i=1,\ldots, q$,
  we let   $\chi_{A_i}(k)$ denote the characteristic function of $A_i$ and we let 
  $\mu_i\ne 0$ be real numbers. An {\it $M_A$-covering function} for a system \eqref{e1} is a sum
    \begin{equation} \label{e2}
   M_A(k)=\sum_{i=1}^q \mu_i \chi_{A_i}(k),
   \end{equation} 
 If all $\mu_i=1$, then  $M_A(k)$ is the usual {\it covering function}:  $w_A(k)=\sum_{i=1}^q  \chi_{A_i}(k) $.
 \end{definition}
  
  Observe that the function $M_A(k)$ is $N_A$-periodic with   $N_A=\lcm(n_1,\ldots,n_q) $. 
  \begin{definition}
  A system \eqref{e1} is called a  {\it covering system} or a {\it cover} (of $\mathbb Z$) if $w_A\ge1$.
   \end{definition}
  Paul  Erd\H{o}s  introduced the concept of the covering system in the 1930s (see, e.g., \cite{E}) and 
  gave a nontrivial example of such a system: $0(2), 0(3), 1(4),
  3(8), 7(12), 23(24)$.
      
     \begin{definition}
    A system \eqref{e1}  is an {\it exact $M$-cover} if $w_A(k)=M$ and an {\it exact cover} if $M=1$. 
 We also say that  $A$ is an  exact $M$-cover of $B$ if $w_A(k)=Mw_B(k)$. 

   \end{definition}
   In what follows,  $  \big(B_m\big)_{m\ge0}=\big(1,-{1\over2}, {1\over6}, 0, -{1\over30},0, {1\over42},\ldots \big)  $
   are the Bernoulli numbers and $B_m(t)$  are the Bernoulli polynomials \eqref{pol-Bm}.
 In 1973 Fraenkel  \cite{Fr} found a connection between  exact covers and  
 $B_m(t)$. Two years later  Porubsk\'y  \cite{P2} extended  it to  exact $M$-covers.
 
    \begin{theorem}  [{\text A. Fraenkel, $M=1$}]   \label{T1}
    A system of $q\ge2$ arithmetic sequences 
    $\{a_i(n_i)\}_{i=1}^q$ is an exact $M$-cover if and only
    if 
   \begin{equation} \label{e3} 
   MB_m= \sum\limits_{i=1}^{q}   n_i^{m-1}  B_m\Big( {a_i\over n_i}\Big)
   \end{equation} 
    holds for  every $m\ge0$.
   \end{theorem}
     Note that for $m=0,1$  the identity  \eqref{e3}  gives us   two well known properties of   exact 
     $M$-covers: 
    \ $  \sum_{i=1}^{q} {1\over n_i}=M$ (see also Corollary \ref{C1}) \ and \ $\sum_{i=1}^{q} {a_i\over n_i}={q-M\over2}  $. 

   Beebee  \cite{B} observed that \eqref{e3}  is the case $x=0$ of a more general identity \eqref{e4}.
   
     \begin{theorem}  [{\text  J. Beebee, $M=1$}]    \label{T2} 
    A system of $q\ge2$ arithmetic sequences 
     $\{a_i(n_i)\}_{i=1}^q$  is an exact $M$-cover if and only if 
   \begin{equation} \label{e4} 
   MB_m(x)=\sum\limits_{i=1}^{q}   n_i^{m-1}  B_m\Big( {x+a_i\over n_i}\Big)
   \end{equation} 
     holds for  every $m\ge0$.
   \end{theorem}
  
   Porubsk\'y \cite{P, P2} used Beebee's approach to extend Theorems \ref{T1} and  \ref{T2} to a  system  \eqref{e1} 
  with  an $M_A$-covering function \eqref{e2}  (and  restrictions $q\ge2, \ N=N_A$, which we omit).
 
    \begin{theorem}        [{\text {\v S}. Porubsk\'y \cite[Th.\,1]{P}}]    \label{T3}
 Let $ q, S \in  \mathbb Z^+$ and   $N=SN_A$. Then
  \eqref{e2} is an $M_A(k)$-covering function  for  the system \eqref{e1} 
  if and only if 
  \begin{equation} \label{e5} 
N^{m-1} \sum\limits_{k=0}^{N-1}  M_A(k)  B_m\Big( {x+k\over N}\Big)
= \sum\limits_{i=1}^{q}   \mu_i n_i^{m-1}  B_m\Big( {x+a_i\over n_i}\Big)
  \end{equation}
  holds for  every $m\ge0$.
  \end{theorem}
  
   We prove that, by itself,  identity \eqref{e5}  is a sum of 
   Raabe identities for arithmetics sequences  $a_i(n_i) $ (see Theorem \ref{T0} and Example \ref{E1}).
   Aside from this result, our other contribution to Theorem \ref{T3} is proving its corollary, Theorem \ref{T4}, where
   we take advantage of $N=SN_A$ rather than $N=N_A$. 
    Theorem \ref{T4} generalizes the results of Beebee and Fraenkel and  is 
      useful in applications to covering systems. In particular, it can be used to check if a system is an exact $M$-cover
      of another system (see Examples \ref{E3} and \ref{E4}).
   In Section \ref{S4} we present Porubsk\'y's proof of Theorem \ref{T3}. In Theorem \ref{T7} of that section 
   we derive identity \eqref{e5} again, this time,  from the asymptotic expansion of the Hurwitz  zeta function.
    In  Section \ref{S5} we give a few examples. 
    
     We begin the article by showing  that many known  identities for Bernoulli numbers,
     including those of Stern and  Namias, are  special cases of 
   the multiplication formula proved by Raabe in his 1848 work
  ``The Jacob Bernoulli function" \cite{Ra}. 
   
\section{Multiplication  formula of J. L. Raabe (1848)}  \label{S2}
       
Raabe \cite{Ra} proved his celebrated multiplication formula: 
     \begin{equation} \label{e19}
   T^{m-1} \sum\limits_{l=0}^{T-1} b_m\Big({t+l\over T}\Big) = b_m(t) -{B_m\over m}\Big(T^m-1\Big), \qquad m\ge1,
    \end{equation} 
  for  polynomials  $\big( t, {t^2\over2}-{t\over2}, {t^3\over3}-{t^2\over2}+{t\over6},\ldots \big) $,
       called  {\it the Bernoulli functions} in \cite{Ra},
  \footnote{Raabe's 1851 article  \cite{Ra2}   is often cited instead of his 1848 monograph \cite{Ra}   
 as the earliest place in the literature  where the term  {\it Bernoulli polynomial (function)} appears.}
        \begin{equation} \label{fn-bm}
    b_m(t)= {1\over m} \sum_{s=0}^{m-1}  B_s{m\choose s} t^{m-s}.  
        \end{equation} 
     
Modern Bernoulli polynomials $B_m(t)$ are  linearly related to  $b_m(t)$,
as follows:
        \begin{equation} \label{pol-Bm}
         B_m(t) = \sum_{s=0}^{m}  B_s \, {m\choose s} \,  t^{m-s}  = m\,b_m(t)+B_m.
       \end{equation}      

Polynomials $b_m(t)$  are solutions to Jacob Bernoulli's  problem: find polynomials that sum  
     powers of consecutive integers $ \sum\limits_{j=1}^{t-1} j^{m-1} = b_m(t)$   when $m, t &gt;1$
      \big(\,e.g., $ \sum\limits_{j=1}^{30} j^{7} = b_{8} (31)$\,\big).
   
   To prove \eqref{e19}, Raabe observed that for odd $m$  the property $b_m(1-x)=(-1)^mb_m(x)$, written as 
  a cancellation law
   $b_{2n+1}(1-x)+b_{2n+1}(x)=0$, yields $b_{2n+1}({1\over 2})=0$  and a general cancellation law, that is,
   multiplication formula \eqref{e19}  at $t=0$ and odd $m\ge3$:
 $$\begin{matrix}  b_{2n+1}({1\over T})+b_{2n+1}({2\over T})+\cdots 
 + b_{2n+1}({T-2\over T})+b_{2n+1}({T-1\over T}) =0  .
 \end{matrix}$$
 He then used impressive algebra to obtain  \eqref{e19}  for all $t$ and $m$.
 
 Raabe's identity \eqref{e19}  can be viewed as a recurrence relation for the Bernoulli numbers: 
        \begin{equation} \label{e22}
       B_m= {T^{m-1}   \over 1-T^m} \, \sum\limits_{l=0}^{T-1} \ \widehat b_m\Big({t+l\over T}\Big) - \widehat b_m(t),
        \end{equation} 
       where $\widehat b_m(t)=m\,b_m(t)$, 
     which are sometimes  called {\it Bernoulli polynomials of an older type}.  
     If we take $t=0$  in the last identity, we see that it coincides with  the following  known identities:
    
     \noindent J. Stern \cite{St}, $T=2$:
         $$B_m={1 \over 2(1-2^m)}  \sum\limits_{s=0}^{m-1} B_s {m\choose s} 2^s  \  
        \buildrel \eqref{e22}  \over  \Leftrightarrow \
        B_m= {2^{m-1} \over 1-2^m}  \ \widehat b_m\Big({1\over 2}\Big).  \qquad
       $$ 
    V. Namias \cite{N}, $T=3$:
         $$B_m= {1 \over 3(1-3^m)}  \sum\limits_{s=0}^{m-1} B_s {m\choose s} 3^s(1+2^{m-s})  
         \  \buildrel \eqref{e22} \over \Leftrightarrow \ 
        B_m= {3^{m-1} \over 1-3^m}   \sum\limits_{l=0}^{2} \,  \widehat b_m\Big({l\over 3}\Big). 
        $$ 
   Namias \cite{N} conjectured  (and Deeba, Rodriguez  \cite{DR} proved) that for arbitrary $T$
        $$ B_m=
  {1 \over T(1-T^m)} \sum\limits_{s=0}^{m-1} T^{s} \sum\limits_{l=1}^{T-1}   B_s {m\choose s} l^{m-s}
    \  \buildrel \eqref{e22} \over \Leftrightarrow \   B_m= {T^{m-1}  \over 1-T^m}    
    \sum\limits_{l=0}^{T-1} \,  \widehat b_m  \Big({l\over T}\Big).
 $$
  In \cite{B,P}  these identities \big(with our notation $\widehat b_m(t)$  for the sum 
  $\sum_{s=0}^{m-1}  B_s{m\choose s} t^{m-s} $ \big) were cited as special cases of Fraenkel's identity \eqref{e3}. 
  The latter is the identity \eqref{e5}  in the  recurrence form with
  $t=0$ and  $M_A(k)=M$:
          \begin{equation} \label{e23} 
         B_m \Big( M-\sum\limits_{i=1}^q   n_i^{m-1}\Big) =
                    \sum\limits_{i=1}^{q}   n_i^{m-1}  \ \widehat b_m\Big( {a_i\over n_i}\Big). 
        \end{equation}
  
   \section{Raabe's identity for arithmetic sequences} \label{S3}
   
     When  identity \eqref{e5}  (with $N=T$ and $x=t$) is applied to the sequence $0(1)=\mathbb Z$,  it becomes the multiplication 
     formula \eqref{e19}  written in terms of  modern Bernoulli polynomials  \eqref{pol-Bm}:
    
      \begin{equation} \label{e6}   
      T^{m-1}\sum_{l=0}^{T-1} B_m\Big({t+l\over T}\Big)= B_m(t). 
         \end{equation} 
  A substitution $ t={x+a_i\over n_i} $ in \eqref{e6} yields  Raabe's identity  for an arithmetic sequence $a_i(n_i)$:  
     \begin{equation} \label{e7}
     T^{m-1} \sum_{l=0}^{T-1} B_m\left( {{x+a_i\over n_i}+l \over  T}\right)  = {B_m\left({x+a_i\over n_i}\right) }.
    \end{equation} 
   
    \begin{lemma}  \label{L1}
  Let $A_i=a_i(n_i),\ a_i&lt;n_i$ and  let $ T\in \mathbb Z^+ $, $N=Tn_i $,
  Then for any function $f$
  
 $$ \sum_{k=0}^{N-1}  \chi_{A_i}(k) f(k)=\sum_{l=0}^{T-1} f(a_i+ln_i).$$
 \end{lemma}
  \begin{proof}  
 The set of integers $\{0,1,2,\ldots, N-1\}$ is split into $T$ subsets:
  $$
  \big\{0,1,\ldots,n_i-1;\ \ n_i,\ldots\,,2n_i-1;\ \ \ldots \,; \ \ (T-1)n_i,  \ldots\,, Tn_i-1\big\}.
  $$
 In each of these subsets there is exactly one member of $A_i$.  
 \end{proof}
 
   \begin{corollary} \label{C1}
  Let \eqref{e1} be a cover. Then Lemma  \ref{L1} with  $f=1$ and $N=SN_A, \, S\in \mathbb Z^+$ yields
 $$  \sum\limits_{k=0}^{N-1}  \chi_{A_i}(k)=T={N\over n_i}.  \quad  {\text Thus} \quad  
 {1\over N} \sum\limits_{k=0}^{N-1}  w_{A}(k)  =\sum\limits_{i=1}^q{1\over n_i} 
 \quad  \hbox{ and } \quad \sum_{i=1}^q{1\over n_i} \ge1.
 $$
 \end{corollary}
  
   \begin{theorem}      \label{T0}
     Identity \eqref{e5}  for a system $A$ in \eqref{e1} with an $M_A$-covering function \eqref{e2}
     is  the sum  of $q$ identities \eqref{e7}  multiplied by $\mu_i$.
     \end{theorem}
     \begin{proof}
   With $f(k)=B_m\big({x+k\over N}\big)$ and $N=Tn_i$ in Lemma \ref{L1}, Raabe's identity \eqref{e7} for $a_i(n_i)$  
   admits the following form: 
  \begin{equation} \label{e8}
 N^{m-1}\sum_{k=0}^{N-1}  \chi_{A_i}(k) B_m\left({x+k\over N}\right)=
  n_i^{m-1} B_m\left({x+a_i\over n_i}\right)  .  
    \end{equation} 
   Now if $N=SN_A$,  then the sum  of $q$ identities \eqref{e8}  multiplied by $\mu_i$ is
   \begin{equation} \label{e9}
 \sum\limits_{i=1}^{q}   \mu_i 
   N^{m-1}\sum_{k=0}^{N-1}  \chi_{A_i}(k) B_m\left({x+k\over N}\right)
  =\sum\limits_{i=1}^{q}   \mu_i  n_i^{m-1}{B_m\left({x+a_i\over n_i}\right) } , 
  \end{equation} 
  which becomes \eqref{e5}  after we change the order of summation in the left side of \eqref{e9}. 
  \end{proof}
   
  \begin{theorem}            [{\text Corollary to Theorem \ref{T3}}]    \label{T4}
  A covering system $A=\{a_i(\, n_i)\}_{i=1}^q$ 
   is an exact $M$-cover of a covering system $B=\{h_j(\, r_j)\}_{j=1}^p$  if and only if 
   \begin{equation} \label{e10}
    M \sum\limits_{j=1}^{p}     r_j^{m-1}{B_m\left({x+h_j\over r_j}\right) } =
    \sum\limits_{i=1}^{q}     n_i^{m-1}{B_m\left({x+a_i\over n_i}\right) }   
   \end{equation} 
   holds for  every $m\ge 0$.
  \end{theorem}
    \begin{proof} Let  $N= \lcm (n_1,\ldots,n_q;\, r_1,\ldots,r_p) =S_1 N_A=S_2N_B$. 
   We write \eqref{e5} for systems $A$ and $B$. By the assumption, $w_A(k)=Mw_B(k)$, i.e.,
   the left side of \eqref{e5} for $A$ equals that for $B$ multiplied by $M$.  
   Hence, the right sides of \eqref{e5} for $A$ and $B$ satisfy  Eq.~\eqref{e10}. 
     \end{proof}
  
        \begin{theorem} [{\text Special case of Theorem \ref{T4}}] \label{T5}
   A covering system  $A=\{a_i(\, n_i)\}_{i=1}^q$ 
    is an exact $M$-cover of an arithmetic sequence $h(r)$ if and only if for  every $m\ge0$:
   \begin{equation} \label{e11}
    M r^{m-1} B_m \left({x+h\over r}\right)= \sum\limits_{i=1}^{q} n_i^{m-1}{B_m\left({x+a_i\over n_i}\right) }. 
     \end{equation} 
     \end{theorem}
   
 \begin{remark} Theorem \ref{T5} tells  us that  writing Raabe's identity \eqref{e7} for $a_i (n_i)$ for all $m\ge0$
       is equivalent to exactly covering  $a_i (n_i)$ by  the system
    $\{(a_i + ln_i) (n_iT), \ l=0, \ldots, T-1 \}$. 
    \end{remark} 
    
  \section{Proof of Theorem \ref{T3}}  \label{S4}
     
  \begin{lemma} \label{L2}  Let  a function $M(z)$ be $N$-periodic and  a function $g(z)$ satisfy
 $ \sum\limits_{z=0}^\infty   |g(z)|&lt; \infty$. 
 Then the summation in the series $\sum M(z)\,g(z)$ can be rearranged as follows:
     $$
     \sum_{z=0}^\infty  M(z) \, g(z) =\sum_{k=0}^{N-1} \sum_{l=0}^\infty  M(k+lN) g(k+lN)
    =\sum_{k=0}^{N-1} M(k) \sum_{l=0}^\infty   g(k+lN).
    $$
      \end{lemma} 
  
  \begin{corollary} \label{C3}
   Let \eqref{e2} be an $M_A(k)$-covering function  for the system \eqref{e1} and
  let   $N=SN_A$,  $S\in \mathbb Z^+$. In addition,  let  $|y|&lt;1$. Then    
    \begin{equation} \label{e12}
  \sum_{k=0}^{N-1} M_A(k) {y^k\over 1-y^N}= \sum_{i=1}^q\mu_i  {y^{a_i}\over 1-y^{n_i} }.
  \end{equation} 
 \end{corollary}

\begin{proof}
   Because $M_A(z)$ is $N$-periodic, using geometric series expansions, the fact that $a_i&lt;n_i$, and Lemma \ref{L2}, 
   we conclude that
    \begin{equation} \label{e13}
   \sum_{i=1}^q \mu_i  {y^{\,a_i}\over 1-y^{n_i} }= \sum_{i=1}^q\mu_i    \sum_{l=0}^\infty  y^{\,a_i+l n_i} 
   = \sum_{i=1}^q\mu_i    \sum_{z=0}^\infty  \chi_{A_i}(z) \,y^{\,z}  =  \sum_{z=0}^\infty  M_A(z)\, y^{\,z} 
     \end{equation} 
     $$
    = \sum_{k=0}^{N-1}M_A(k)  \sum_{l=0}^\infty  y^{\,k+l N} = \sum_{k=0}^{N-1}   M_A(k) {y^{\,k}\over 1-y^N}. 
    \eqno \square $$
\end{proof}

   
  \begin{corollary} [{\text {\v S}. Porubsk\'y \cite{P2}}]  \label{C4}  
 Let  $N=SN_A$, $S\in \mathbb Z^+$. Then \eqref{e2} is an $M_A(k)$-covering function  
 for the system \eqref{e1} if and only if Eq.~\eqref{e12} holds for   any $y$ with $|y|&lt;1$.  
 \end{corollary}
  We recall Euler's generating function of polynomials $ B_m(t)$:
   \begin{equation} \label{e14}
 {se^{st}\over e^s-1}= \sum\limits_{m=0}^{\infty} {B_m(t)\over m!} \, s^m .
  \end{equation} 
 
      \begin{proof} [Proof of Theorem \ref{T3} {\text (J. Beebee, \v S. Porubsk\'y)}]
\leavevmode
\vphantom{a} \\
\medskip
\noindent $(\Leftarrow)$:    
We assume that Eq.~\eqref{e5} holds for every $m\ge 0$. 
  As in \cite{B,P}, we multiply \eqref{e5}  by ${s^m / m!}$ and sum the result over   $m$: 
 
      $$ \sum\limits_{k=0}^{N-1}   M_A(k)   \sum\limits_{m=0}^{\infty} {(sN)^m\over N\ m!}  B_m\Big( {x+k\over N}\Big) 
      = \sum\limits_{i=1}^{q}   \mu_i
 \sum\limits_{m=0}^{\infty} {(sn_i)^m\over n_i \ m!}  B_m\Big({x+a_i\over n_i}\Big).
  $$
 By Eq.~\eqref{e14} this is equivalent to
   $$ \sum\limits_{k=0}^{N-1}   M_A(k)   {(sN)\over N}  {e^{sN(x+k)/ N} \over e^{sN} -1}
      = \sum\limits_{i=1}^{q}   \mu_i
 {(sn_i)\over n_i } {e^{sn_i(x+a_i)/ n_i} \over e^{sn_i} -1},  
 \eqno $$
which, with $e^s=y$,  becomes \eqref{e12}. Thus $M_A$ satisfies \eqref{e2} by  Corollary \ref{C4}.

\medskip

\noindent $(\Rightarrow)$:   Let $M_A$ satisfy Eq.~\eqref{e2}. Starting with \eqref{e12} we 
 reverse  steps in the above argument and obtain Eq.~\eqref{e5} for every $m\ge 0$. 
\end{proof}

 The authors of the proofs of Theorems \ref{T2} and \ref{T3}  use the generating function of integers employed in \eqref{e13} 
 of Corollary \ref{C3} to Lemma \ref{L2} together with Euler's generating function  \eqref{e14}.
  
  Next we derive identity \eqref{e5} from the asymptotic expansion of the Hurwitz 
    zeta function using  just Lemma \ref{L2}. 
 
        \begin{theorem}  \label{T7}
  Identity \eqref{e5} holds for  any system $A$ in \eqref{e1} with an $M_A$-covering function \eqref{e2}.
  \end{theorem} 
  \begin {proof} 
   By  Euler-Maclaurin summation (see, e.g., \cite[Ex.\,3, \S467]{F}, where $L+x=a$)
 \footnote{ Fichtenholtz \cite{F} uses the notation $\beta_n$ for the Bernoulli numbers and $B_n$ 
 for the absolute values of nonzero Bernoulli numbers. More precisely, 
  $\big(\beta_n\big)_{n\ge1}=\big(-{1\over2}, {1\over6}, 0, -{1\over30},0, {1\over42},\ldots\big)$ and 
  $\big(B_{n}\big)_{n\ge1} =\big({1\over6}, {1\over30}, {1\over42}, {1\over30},{5\over66},\ldots\big)$.
 }
     the asymptotic series  of an $L$-th tail of the Hurwitz zeta function at $2$ in powers of $1/(L+x)$ is
       \begin{equation} \label{L-tale}
                      \sum\limits_{l=0}^\infty{1 \over (L+x+l)^2}   
              \sim  \sum_{k=0}^{\infty}   {(-1)^{k} B_{k}\over(L+x)^{k+1}} .
    \end{equation} 
   The asymptotic series composed with the converging series yields the  series in powers of ${1/L}$:
$$ 
 \sum\limits_{l=0}^\infty{1 \over (L+x+l)^2}  \sim
 \sum_{k=0}^{\infty}{(-1)^{k} B_{k}\over(L+x)^{k+1}}=\sum\limits_{k=0}^{\infty} {(-1)^{k} B_{k}\over L^{k+1}} 
 \sum\limits_{j=0}^{\infty} {k+j \choose j}{(-x)^j\over L^j} 
 $$ 
  \begin{equation} \label{e15}  \buildrel k+j=m \over
 = \ \sum\limits_{m=0}^{\infty} {(-1)^{m}\over L^{m+1}}\sum\limits_{k=0}^{m} B_{k}\,{m \choose m-k} x^{m-k}       
   \,  \buildrel \eqref{pol-Bm} \over  =  \,  \sum_{m=0}^{\infty}   {(-1)^m\over L^{m+1}}  \,   B_{m}(x) . 
      \end{equation} 
      Let $N=SN_A, \ S\in \mathbb Z^+$. 
   Lemma \ref{L2} and Eq.~\eqref{e15}  with $L$ replaced by  $L/N$ and $x$ by $ (x+k)/N$ give us
    \begin{equation} \label{e16}   
  \sum\limits_{z=0}^\infty{M_A(z) \over (L+x+z)^2} 
  = \sum\limits_{k=0}^{N-1}M_A(k)\sum\limits_{l=0}^\infty{1 \over (L+x+k+l N)^2}
     \end{equation} 
   $$ 
   = \sum\limits_{k=0}^{N-1} {M_A(k) \over N^2}  \sum\limits_{l=0}^\infty{1\over ({L\over N}+{x+k\over N}+l )^2}     
      \sim \sum_{m=0}^{\infty}   
    {(-1)^m\over L^{m+1}}  \sum\limits_{k=0}^{N-1}  M_A(k)N^{m-1} B_{m}\Big({x+k\over N}\Big) .
  $$
  Similarly, Eq.~\eqref{e15} with $L$ replaced by   $L/n_i$ and $x$ by   $ (x+a_i)/n_i$  gives us
     \begin{equation} \label{e17}  
  \sum\limits_{z=0}^\infty{ \sum_{i=1}^q \mu_i \chi_{A_i}(z) \over (L+x+z)^2} =
  \sum\limits_{i=1}^q \mu_i  \sum\limits_{l=0}^\infty{1 \over (L+x+a_i+l n_i )^2}
     \end{equation} 
  $$= \sum\limits_{i=1}^q {\mu_i\over n_i^2} \sum\limits_{l=0}^\infty{1 \over  ({L\over n^i}+{x+a_i\over n_i}+ l)^2}
       \sim \sum_{m=0}^{\infty}   {(-1)^m \over L^{m+1}} 
       \sum\limits_{i=1}^q  \mu_i n_i^{m-1} B_{m}\left({x+a_i\over n_i}\right) . 
       $$  
    Now Eqs.~\eqref{e16} and \eqref{e17} yield  Eq.~\eqref{e5}, because the asymptotic series is unique.   
     \end{proof}
     We note that, unlike Theorem \ref{T3}, 
     steps in the last theorem cannot be reversed  because functions can share asymptotic series. 
  
   We conclude the section with the famous proof of a relation for moduli  of the exact covers 
   found by Mirsky and  Newman, and independently, by Davenport and Rado (see, e.g., \cite{E2}).
     
        \begin{theorem}  
   [\text H. Davenport, L. Mirsky, D. Newman, R. Rado]  \label{T8}       
 Let a  system \eqref{e1} be an exact M-cover and let its moduli $n_i$ satisfy   $1&lt; n_1\le n_2\le \cdots \le n_{q-1}\le n_q $. 
 Then $n_{q-1}=n_q$.
 \footnote{ Further results for  covering systems in this direction can be found in \cite{GS, S} and in
 \textquotedblleft Covering systems and periodic arithmetical functions"  
 (a talk given by Z. W. Sun  at UIUC on April 13, 2006).
 }
 \end{theorem}
  
 \begin{proof}
 In Eq.~\eqref{e13} of Corollary \ref{C3} we take $\mu_i=1,\ i=1,\ldots,q$ and $M_A(z)=M$   to have 
    \begin{equation} \label{e18}
 {M\over 1-y}  = \sum\limits_{i=1}^{q-1}   {y^{a_i}\over 1-y^{n_i} }+ {y^{a_q}\over 1-y^{n_q} }.
    \end{equation} 
The assumption that $n_{q-1}&lt;n_q$  would contradict Eq.~\eqref{e18}  when $y\to e^{2\pi i / n_q}$:
 $$   \lim_{y\to e^{2\pi i / n_q}}
    \left(  {M\over 1-y}- \sum\limits_{i=1}^{q-1}    {y^{a_i}\over 1-y^{n_i} }\right) \
    \ne \lim_{y\to e^{2\pi i/n_q}} {y^{a_q}\over 1-y^{n_q} } = \infty .  $$
    \end{proof}  
    
       \section{Examples }  \label{S5}
              
      Identity \eqref{e5} in the  recurrence form with $t=0$ and $ M_A(k)=w_A(k)$   was given   in  {\cite[Cor.\,1]{P}}:
       \begin{equation} \label{e24}  
     B_m   \sum\limits_{i=1}^q \Big( {N^{m}\over n_i} -  n_i^{m-1} \Big)
     =  \sum\limits_{i=1}^{q}   n_i^{m-1} \, \widehat b_m\Big( {a_i\over n_i}\Big)  - 
     N^{m-1} \sum\limits_{k=0}^{N-1} w_A(k) \, \widehat b_m\Big( {k\over N}\Big). 
          \end{equation} 
        The term  $ \sum_{i=1}^q  {N\over n_i}$ in the left side of Eq.~\eqref{e24} replaced 
        $ \sum_{k=0}^{N-1} w_A(k)$   (see Corollary \ref{C1}).
 
      \begin{example} \label{E1}
        Identity \eqref{e24} applied to the system  $A=\{0(2), 0(3), 1(6), 5(6)\}$  with $N=6$ is 
       $$
        \hbox{$ B_m  \Big(2^{m-1} +3^{m-1} -   5\cdot 6^{m-1} \Big)   = 6^{m-1}\,
   \Big(\,\widehat b_m\big({2\over6}\big)  + \widehat b_m\big({3\over6}\big) +\widehat b_m\big({4\over6}\big)\Big).$}
      \eqno \hbox{\cite[Cor.\,3]{P}} 
      $$
      Since  $ \widehat b_m(t) =B_m(t)-B_m$,
      the above identity is equivalent to 
\begin{equation} \label{e25}
    \begin{matrix} B_m\Big(   2^{m-1} +3^{m-1} - 2\cdot  6^{m-1}\Big) =
  6^{m-1}\Big( B_m\big({2\over6}\big)  + B_m\big({3\over6}\big) +B_m\big({4\over6}\big)\Big). 
   \end{matrix} 
     \end{equation} 
  The latter is the  sum of Raabe identities \eqref{e7} for $0(2)$ with $T={6\over2}$ and for $0(3)$ with $T={6\over3}$:
       $$ \begin{matrix}
     \quad  6^{m-1}\Big( B_m+B_m\big({2\over6}\big)  + B_m\big({4\over6}\big)\Big) = 2^{m-1}B_m  
    \quad  \hbox{and} \quad  6^{m-1}\Big( B_m+B_m\big({3\over6}\big)  \Big) = 3^{m-1}B_m . 
    \end{matrix} 
    $$
  Note that in Theorem \ref{T0} identity \eqref{e25} for $A$ is a sum \eqref{e9} of 4 identities  
    \eqref{e7}, but  identity \eqref{e7} applied to the arithmetic sequence 1(6) or 5(6)  yields just  
    $B_m=B_m$.
\end{example}
   
         \begin{example} \label{E2}
   Nevertheless,  we could apply Eq.~\eqref{e24} in the original form \eqref{e5}  to a system $A^\prime$ related 
   to $A$, for example, to find $B_m\big({1/6}\big)$ without knowledge   of $B_m({1/2})$ and $ B_m({1/3})$:
         \begin{equation} \label{e26} \begin{matrix}
      B_m\big({1\over6}\big)=  B_m (1-2^{m-1})(1-  3^{m-1}) / ( 6^{m-1}  \,2) .
      \end{matrix}  \end{equation} 
    According to Theorem \ref{T5}, writing identities \eqref{e25}  is  equivalent to covering  $0(3)$ by  $\{0(6), \, 3(6)\}$
    and $0(2)$ by $\{0(6), \,2(6),\, 4(6)\}$.  Clearly  systems
      $A^\prime=$\{0(6), 2(6),\,4(6),\,0(6),\,3(6),\,1(6),\,5(6)\} and  $A$ have the same covering function:
$w_A(k)=2$ if $ k \equiv \modd{0} {6}$  and $w_A(k)=1$ otherwise. 
   We  apply Eq.~\eqref{e5}  to  $A^\prime$ with  $  M_{A^\prime}=w_A$, $N=N_A=6$ and $t=0$ to get
  $$
      6^{m-1} \sum\limits_{k=0}^5   B_m\big( {k\over 6}\big) + 6^{m-1}  B_m(0) 
     =   2^{m-1}B_m(0)  +3^{m-1}  B_m(0) + 6^{m-1}  \Big( B_m\big( {1\over 6}\big)+  B_m\big( {5\over 6}\big)\Big)  
$$
 and use  $B_m\big( {5\over 6}\big)=(-1)^m B_m\big( {1\over 6}\big)$   and Raabe's identity
  $6^{m-1} \sum\limits_{k=0}^5   B_m\big( {k\over 6}\big) =B_m$ 
 to  prove Eq.~\eqref{e26}.
  \end{example} 
The next two examples show  how their exact covers   were constructed.
      
  \begin{example} \label{E3}
     Choi  constructed an exact $2$-cover  that is not the union of  two exact covers:
   $1(2), 0(3), 1(3), 2(6), 0(10), 4(10), 6(10),   8(10),  2(15), 5(30),  11(30), 12(30), 22(30), 23(30), 29(30)$
   (see, e.g., \cite[p.\,156]{P}).
  To confirm that this system is an exact $2$-cover,
  we split  its sequences  $0(3), 1(3), 2(15)$  into odd 
  and even terms and show that  the modified  cover is the union of  an exact $2$-cover  $B \cup  C$ of the  
  sequence $1(2)$  and  an exact $2$-cover $D \cup  E$ of the sequence $0(2)$: 
\begin{align*}
  B &amp;=1(2); &amp; \quad C&amp;= \big\{1(6), 3(6), 5(30), 11(30), 17(30), 23(30), 29(30) \big\} ;   \\
   D  &amp;= \big\{  0(6), 2(6), 4(6) \big\}; &amp; \quad E &amp;= \big\{ 0(10), 4(10), 6(10), 8(10), 2(30), 12(30), 22(30) \big\} .
\end{align*}
 Since $B$ and $D$ are already  exact covers of 1(2) and  0(2) respectively, we use Theorem \ref{T5}  
 with $M=1, \ x=0 $ to show that  $C$ and $E$ are such covers.  
   
   First,  in Theorem \ref{T5} we let the sequence $h(r)$ be $1(2)$ and the system $A$ be $C$. 
   In that case, the right side of \eqref{e11} in Theorem \ref{T5} is  
\begin{multline*}
     6^{m-1} B_m\big({1\over6}\big)  +6^{m-1} B_m\big({3\over6}\big)   \ + \
    6^{m-1} 5^{m-1} \sum\limits_{l=0}^{5-1} B_m\big({5+6l\over30}\big)  
   \\ 
     =\ 6^{m-1} B_m\big({1\over6}\big)  +6^{m-1} B_m\big({3\over6}\big)  + 6^{m-1}  B_m\big({5\over6}\big)
  \ = \ 2^{m-1}  B_m\big({1\over2}\big), 
\end{multline*}
which is the left side of \eqref{e11}. We  have applied Raabe's identity \eqref{e7} 
with $T=5, \ x={5\over6}$  to the sum in the first row and  with   $T=3, \ x={1\over2}$ in the second row.
    
 Now, similarly, in Theorem \ref{T5} we take  $h(r)= 0(2) $, $A=E$ and with the help of \eqref{e7} confirm  \eqref{e11}:
\begin{align*}
  10^{m-1} \sum\limits_{l=0,2,3,4} B_m\big({2l\over10}\big)
  + 10^{m-1} 3^{m-1} \sum\limits_{l=0}^{3-1} B_m\big({2+10l\over30}\big) 
  =10^{m-1} \sum\limits_{l=0}^{5-1} B_m\big({l\over5}\big) =2^{m-1} B_m({0\over2}). 
\end{align*} 
       \end{example} 
      
   \begin{example} \label{E4}
    Show that a system
   $ F\cup G =\{2^{i-1}( 2^i)\}_{i=1}^f \cup  \{(j-1)2^{f}( n 2^f)\}_{j=1}^n $   is an exact cover.  
    First, we use \eqref{e6}   with $t=0$ and $T=n$ to see that $G$ is an exact cover of $0(2^f)$: 
       \begin{align} 
          \sum\limits_{i=1}^{f}   (2^i)^{m-1} B_m\big({1\over 2}\big)+
        (2^f)^{m-1}  n^{m-1}\sum\limits_{l=0}^{n-1}   B_m \big({l\over n}\big) 
	\label{e27} \\
      = \sum\limits_{i=1}^{f} (2^i)^{m-1} B_m \big({1\over 2}\big) +  (2^f)^{m-1}  B_m (0) \label{e28}.
     \end{align} 
 Then we apply Theorem \ref{T5}  repeatedly using \eqref{e6} with $t=0$ and $T=2$ to see that  
 $2^{f-1}( 2^f)\cup 0(2^f)$ is an exact cover of $0(2^{f-1})$,   etc.: 
\begin{align*}
     \eqref{e28}=  \ \sum\limits_{i=1}^{f-1}   (2^i)^{m-1} B_m \big( {1\over 2}\big)
    +  (2^{f-1})^{m-1} 2^{m-1}  \Big( B_m \big({0\over 2}\big) +B_m \big({1\over 2}\big) \Big)
\\
     = \ \sum\limits_{i=1}^{f-1}   (2^i)^{m-1} B_m \big({1\over 2}\big)
    +  (2^{f-1})^{m-1}  B_m \big({0\over 2}\big) = \cdots = B_m \big({0\over 2}\big)=B_m.
\end{align*} 
 Note that $\eqref{e28}=B_m$ is 
 $B_m\big( {1\over 2}\big) \sum_{i=1}^{f}   (2^i)^{m-1}  =B_m\big(1-(2^f)^{m-1} \big)$  
 which computes $B_m\big( {1\over 2}\big)$. 
  Now $\eqref{e27} =B_m$ with $B_m(t)=\widehat b_m(t)+B_m$ is the relation given in \cite[Cor.\,4]{P}. 
  The latter was \eqref{e23} applied to the system $F\cup G$ and   multiplied by $2^{f+1} (2^{m-1}-1)$. 
   \end{example} 
    
        When  the covering function $w_A$ of a system $A$ is unknown or complicated, the identity
    \eqref{e5} can still compute and estimate the averages of $w_A(k), kw_A(k), \ldots$  over $N_A$.   
      
      \begin{example} \label{E5} For an odd $n\ge3$, the system 
     \  $F \cup H = \{2^{i-1}( 2^i)\}_{i=1}^{n-1} \ \cup \ \{j 2^{n-1}( n2^{j-1})\}_{j=1}^n$ \   \cite[Ex.\,1,\ p.\,4314]{S} 
       is an example of a {\it distinct cover} of $\mathbb Z$  (all $n_i$ are different),
  \footnote{ {\bf Erd\H{o}s-Selfridge Conjecture} (See e.g., \cite{G}). 
 {\it There are no distinct covers with all $n_1, \ldots , n_q$ odd and greater than one.}
    }
    thus by the result of  Davenport-Mirsky-Newman-Rado  it is not an exact cover. 
    To show that it covers $\mathbb Z$, we  take  $f=n-1$  and observe that $0( n2^{f})= n 2^{f}( n2^{f})$
   and, if $0\le j \le f$, the sequence 
   $j 2^{f}( n2^{j-1})$  covers the sequence $j 2^{f}( n2^{f})$. 
   Hence, system $H=\{j 2^{n-1}( n2^{j-1})\}_{j=1}^n$ covers system
       $ G=\{(j-1)2^{f}( n 2^{f})\}_{j=1}^n $  of Example \ref{E4}, while system $F$ is that of  Example \ref{E4}.
       
  As in  \cite{S}, Corollary \ref{C1} computes  the arithmetic average of  
  $w_{F \cup H}(k)$ with   $N=N_{F \cup H}=2^{n-1}n$:    
      \begin{equation} \label{e29} 
     {1\over N} \sum\limits_{k=0}^{N-1}  w_{F \cup H}(k) =\sum\limits_{i=1}^{n-1}{1\over 2^i} 
     + \sum\limits_{j=1}^{n}{1\over n2^{j-1}} = 1+{2^n-n-1\over 2^{n-1} \,n} &lt;1+{2\over n}.
   \end{equation} 
  It shows  that for large $n$ the system $F \cup H$ covers  integers with fewer overlaps.
 
      Note that the system $H=\{h_j( r_j)\}_{j=1}^n=\{j 2^{n-1}( n2^{j-1})\}_{j=1}^n$ is not in the form of \eqref{e1}.  
   In order to apply  \eqref{e5}  with $m\ge1$ to the system $F\cup H$, we define  
   $h_j^{\,\prime}=h_j - k_j r_j $ with $k_j$  so that  $0\le h_j^{\,\prime} &lt; r_j,$ $ j=1, \ldots, n $ 
   and write $H=\{h_j^{\,\prime}(r_j)\}_{j=1}^{n}$,  where  $h_n^{\,\prime}=0$. 
   
   Identity \eqref{e5} at $m=1$  computes the arithmetic average of $ k w_{A}(k) $ for any system \eqref{e1}: 
    \begin{equation} \label{e30} 
  {1\over N} \sum\limits_{k=0}^{N-1}  k w_{A}(k) ={N\over2} \sum\limits_{i=1}^q{1\over n_i} 
  +\sum\limits_{i=1}^q{a_i\over n_i} -{q\over2}.
      \end{equation}  
  Now \eqref{e30} and \eqref{e29}  estimate the arithmetic average of $ k w_{F \cup H}(k) $ for our system:  
 $$
  {1\over N} \sum\limits_{k=0}^{N-1}  k\, w_{F \cup H}(k)  ={N\over2} \sum\limits_{i=1}^q{1\over n_i} +
  \sum\limits_{i=1}^{n-1}{1\over 2} + \sum\limits_{j=1}^{n}{h_j^{\,\prime}\over r_j} -{2n-1\over2}
     &lt;  {N-1\over2} +{N\over n}. 
 $$
For an exact cover  this average would be $ {N-1\over2}$.
  \end{example} 
 
 \section{Acknowledgments} We thank Peter Andrews, Bruce Reznick, John Wetzel,  Daniel Klovsky, and the referee 
 for  their help and valuable comments.
  
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 \end{thebibliography}
  
  \bigskip \hrule \bigskip
\noindent {\it 2010 Mathematics Subject Classification}:  Primary 11B25;
  Secondary 11B68, 11M35. 
\par 
\noindent {\it Keywords}: covering system and function, Bernoulli polynomial, multiplication formula.

\bigskip \hrule \bigskip

\noindent (Concerned with sequences 
\seqnum{A027641} and
\seqnum{A027642}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received March 28 2019;
revised versions received  April 2 2019; January 31 2020; February 5 2020; 
February 11 2020.
Published in {\it Journal of Integer Sequences}, February 20 2020.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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